# CLEP College Algebra/Sequences and Series

In mathematics, it is important to find patterns. That is what mathematicians do almost everyday of their lives. How they determine patterns is different depending on the type of mathematics they work with. For college algebra, determining patterns is part of the curriculum. The problem below demonstrates one way we determine these patterns. Keep in mind that the problem below is an Exploration and most likely not representative of the types of problems you may see in the CLEP College Algebra exam.
${\text{Z}}$
${\text{AA}}$
${\text{AB}},\,{\text{AC}},\,\ldots ,\,{\text{BA}},\,{\text{BB}},\ldots$
${\text{ZZ}}$
${\text{A}}$
${\text{HY}}$
$26$
Concatenating the letter ${\text{A}}$ to ${\text{A}}$ will give us a new letter, ${\text{AA}}$. Adding this to the "Old Alphabet" obviously does not make it the old set anymore, so let us define the alphabet in which you add new letters to the old one the "New Alphabet" How many of the new letters are in this set? Well, each letter from "Old Alphabet" must get another additional letter to make the "New Alphabet" Therefore, for each letter, $26$ letters are added to the old one (i.e. the letter "A" will have letters "A", "B", "C",...,"Z" concatenated to "A", so 26 more letters are added to each letter of the "Old Alphabet"). Since each letter is used, exactly $26^{2}{\text{ new letters}}$ are created in the "New Alphabet" The figure below may help demonstrate this new fact.
${26{\begin{cases}\overbrace {AA,\,AB,\,AC,\ldots ,\,AY,\,AZ} ^{26}\\\overbrace {BA,\,BB,\,BC,\ldots ,\,BY,\,BZ} ^{26}\\{\text{ }}\qquad \qquad \quad \;\;\vdots \\\overbrace {YA,\,YB,\,YC,\ldots ,\,YY,\,YZ} ^{26}\\\overbrace {ZA,\,ZB,\,ZC,\ldots ,\,ZY,\,ZZ} ^{26}\end{cases}}}\!$
The left brace tells us how many letters are in the "Old Alphabet"; the top brace for each "row" tells us how many new letters are created per letter of the alphabet. Since 26 letters are created for each letter of the "Old Alphabet," add every new letter created into this subset of the "New Alphabet." Since there are $26$ rows, and each row creates $26$ new letters, ${\underbrace {26+26+26+...+26+26} _{26}=26\cdot 26=26^{2}=676{\text{ letters in the subset of the New Alphabet.}}}$
Remember that there are still 26 other letters in the "old alphabet" not included in the figure above. As such, add 26 to the answer above to get the total number of letters in the New Alphabet.
Given that we know the total number of letters in the "New Alphabet," we can find out what position letter HY is in this subset of the "New Alphabet." First, the letter position of H is $8$. Second, the letter position of Y is $25$. Since the above figure shows a table, we can find HY by looking up that "coordinate." Since the position of HY contains all the terms in that given area, we can multiply the two values to get the area (i.e. it is a rectangle that contains all those letters). This is not the final answer, however. Realize that we excluded some $7+26$ other letters by multiplying in that block and subset; ergo, the position of the letter ${\text{HY}}$ is
$8\cdot 25+(7+26)=233.$
$26$
Concatenating the letter ${\text{A}}$ to ${\text{A}}$ will give us a new letter, ${\text{AA}}$. Adding this to the "Old Alphabet" obviously does not make it the old set anymore, so let us define the alphabet in which you add new letters to the old one the "New Alphabet" How many of the new letters are in this set? Well, each letter from "Old Alphabet" must get another additional letter to make the "New Alphabet" Therefore, for each letter, $26$ letters are added to the old one (i.e. the letter "A" will have letters "A", "B", "C",...,"Z" concatenated to "A", so 26 more letters are added to each letter of the "Old Alphabet"). Since each letter is used, exactly $26^{2}{\text{ new letters}}$ are created in the "New Alphabet" The figure below may help demonstrate this new fact.
${26{\begin{cases}\overbrace {AA,\,AB,\,AC,\ldots ,\,AY,\,AZ} ^{26}\\\overbrace {BA,\,BB,\,BC,\ldots ,\,BY,\,BZ} ^{26}\\{\text{ }}\qquad \qquad \quad \;\;\vdots \\\overbrace {YA,\,YB,\,YC,\ldots ,\,YY,\,YZ} ^{26}\\\overbrace {ZA,\,ZB,\,ZC,\ldots ,\,ZY,\,ZZ} ^{26}\end{cases}}}\!$
The left brace tells us how many letters are in the "Old Alphabet"; the top brace for each "row" tells us how many new letters are created per letter of the alphabet. Since 26 letters are created for each letter of the "Old Alphabet," add every new letter created into this subset of the "New Alphabet." Since there are $26$ rows, and each row creates $26$ new letters, ${\underbrace {26+26+26+...+26+26} _{26}=26\cdot 26=26^{2}=676{\text{ letters in the subset of the New Alphabet.}}}$
Remember that there are still 26 other letters in the "old alphabet" not included in the figure above. As such, add 26 to the answer above to get the total number of letters in the New Alphabet.
Given that we know the total number of letters in the "New Alphabet," we can find out what position letter HY is in this subset of the "New Alphabet." First, the letter position of H is $8$. Second, the letter position of Y is $25$. Since the above figure shows a table, we can find HY by looking up that "coordinate." Since the position of HY contains all the terms in that given area, we can multiply the two values to get the area (i.e. it is a rectangle that contains all those letters). This is not the final answer, however. Realize that we excluded some $7+26$ other letters by multiplying in that block and subset; ergo, the position of the letter ${\text{HY}}$ is
$8\cdot 25+(7+26)=233.$
What were we doing in the problem above? Essentially, we were simply trying to find what position some "term" is in. Does it not intrigue you to see math try to find a position of some "term" in a list? We have problems like these as mathematicians because the patterns underlying a "list" of numbers can help us determine new facts of mathematics. After all, what were we doing when using functions? We were trying to find a number using a pattern (the function). Unlike the previous sections, however, we were not given a formula. Luckily enough, it is not difficult to make a formula for a given "list" of numbers. Before we dive into these new problems. It helps to establish definitions.
A sequence is a list of elements, such as numbers, figures, or letters, that is generally written in some pattern.
A term is an item found within a sequence.
There is a little disclaimer to get out of the way before we try to solve some problems. First, a sequence can have no pattern. However, for our purposes, we will not count any list of numbers in which no pattern exists. Second, even in simple sequences, numbers of any kind can be named if they follow after a sequence rule. For example, here's a sequence in which the rule is to list prime numbers: ${\left\{{\mathit {2}},{\mathit {3}},{\mathit {5}},{\mathit {7}},{\mathit {11}},\ldots \right\}}$ If you were like most people, you would probably name the prime numbers in order. However, you could perhaps finish the sequence like this: ${\left\{{\mathit {2}},{\mathit {3}},{\mathit {5}},{\mathit {7}},{\mathit {11}},{\mathit {13}},{\mathit {23}},{\mathit {2011}},{\mathit {(2^{82,589,933}-1)}},\ldots \right\}}$ For our purpose of standardization, we will follow a pattern by stating what number you must find first or identify the pattern.
Let's begin exploring the world that is Sequences and Series.

## Sequences

As you already know, a sequence is a list of objects that generally follow a pattern. However, the type of pattern that is described will classify sequences into either arithmetic sequences or geometric sequences. Each will be explored in depth within the next sections.

### Arithmetic Sequences

An index is the location of a term within a sequence, usually denoted by $i$ or $k$.
An arithmetic sequence is a sequence ${\left\{x_{1},x_{2},\ldots ,x_{k},x_{k+1},\ldots \right\}}$ in which an added real number $d$, called the difference, is added to each successive term, except the first term, $x_{1}$, such that the sequence forms ${\left\{x_{1},x_{1}+d,x_{1}+2d,x_{1}+3d,\ldots \right\}}$ in a one-to-one correspondence.
An example could perhaps help you figure with the formal definition above: The sequence ${\left\{2,4,6,8,10,\ldots \right\}}$ has a one-to-one correspondence with the general sequence ${\left\{x_{1},x_{2},x_{3},x_{4},x_{5},\ldots \right\}}$ since the first term $x_{1}=2$, second term $x_{2}=4$, third term $x_{3}=6$, and so on. The difference is the amount added to each previous term to get the new term. For $x_{2}$, $x_{1}+d=x_{2}$ or $2+d=4$. Solving for $d$ is the difference of the two terms. In this example, the difference is $d=2$. This is how we define an arithmetic sequence.

#### Recursive Formula

Often times, we want to generate a sequence using a formula (we are mathematicians, after all, and we like to study sequences to see if there are any general patterns). If we want to find $x_{i}$, we may use the following formula:
$x_{i}=x_{i-1}+d.$
However, the above formula could describe any sequence that has that general pattern. To fix this, we need to describe the first term as well when using the formula above. There are two ways to describe this formula:
* We have it horizontally deliniated: $x_{1}=a;x_{i}=x_{i-1}+d$.
* We have it vertically deliniated: ${\begin{cases}x_{1}=a\\x_{i}=x_{i-1}+d\end{cases}}$
To save space, we will horizontally deliniate formulas for arithmetic sequences in this WikiBook.
Formulas in which the first term is identified along with an equation in which the previous term is added by $d$ to get the next term is called a recursive formula.
A recursive formula is a formula that describes how to get a term or many terms of a sequence by stating the starting value or values and each previous term or terms and adding difference $d$ to each subsequent term.
Given an initial value $a$ and constant difference $d$, the $i$th term in an arithmetic sequence is given by the previous $(i-1)$th term of the sequence:
    $x_{1}=a;x_{i}=x_{i-1}+d$ OR
    ${\begin{cases}x_{1}=a\\x_{i}=x_{i-1}+d\end{cases}}$
* Example 1.1.1.a: Find the $7$th term to the arithmetic sequence ${\left\{19,18.25,17.5,16.75,16,\ldots \right\}}$
* To determine the next term, we first need to find the difference $d$. Note that an arithmetic sequence will have the next term add $d$ to the previous term. Since that is how the arithmetic sequence works, $19+d=18.25$ is a valid way to find the difference between two terms. Solving for $d$, we find the constant difference is $d=-{3 \over 4}=-0.75$. Since the difference is the same for each given term in the sequence, we can find the 7th term by adding $d=-0.75$ to $16$, which gives us $16-0.75=15.25$. Finally, add $d$ to the next term to get the final answer:
* $15.25-0.75=14.5.$
There are many reasons why it is more important to have a recursive formula. It is not always slow; it may be easier to understand. The next example shows why this is exactly true
* Example 1.1.1.b: Write the recursive formula to the sequence ${\left\{1,1,2,3,5,7,13,\ldots \right\}}$.
* Many of you will perhaps know this famous pattern as the Fibonacci Sequence. For those of you who do not know this sequence, the way we determine the next term is by using the previous terms and adding them together. In our notation, we would say that the term $a$ at index $i$, $a_{i}$ is equivalent to $a_{i-2}+a_{i-1}$. Remember, however, we are not done. If a mathematician saw the sequence ${\left\{2,4,6,10,16,26,42,\ldots \right\}}$, he (or she) would determine that $a_{i}=a_{i-2}+a_{i-1}$ also describes that sequence. Therefore, we must list the first two terms because listing only the first term would not allow us to get the next term. This means our final answer is
* $a_{1}=1,a_{2}=1;a_{i}=a_{i-2}+a_{i-1}.$
Note that Example 1.1.b is not an example of an arithmetic sequence. Your next exploration will be to determine why this is true. Along with that, you will use your critical thinking skills to argue for or against something in the explorations after that one.
$x$
$i$
$d$
$d$
$a_{1}=1,a_{2}=1;a_{i}=a_{i-2}+a_{i-1}.$
$d$
$i$
$a_{i-1}$
$d$
$a_{1}=1;a_{i}=a_{i-1}+d$
$x$
$i$
$d$
$d$
$a_{1}=1,a_{2}=1;a_{i}=a_{i-2}+a_{i-1}.$
$d$
$i$
$a_{i-1}$
$d$
$a_{1}=1;a_{i}=a_{i-1}+d$
$x_{1}=a$
$x_{2}=b$
$a\neq b$
* Argue either for OR against the claim that the recursive formula $x_{1}=a,x_{2}=b;x_{i}=x_{i-2}+d$ is arithmetic and recursive.
   * If you disagree that $x_{1}=a,x_{2}=b;x_{i}=x_{i-2}+d$ is arithmetic and recursive, write a formula that is arithmetic and recursive.
   * If you agree that $x_{1}=a,x_{2}=b;x_{i}=x_{i-2}+d$ is arithmetic and recursive, explain why it is so.
Finally, if $x_{1}=a$, $x_{2}=b$, and $a=b$,
* argue either for OR against the idea that the formula $x_{1}=a,x_{2}=b;x_{i}=x_{i-2}+d$ is arithmetic and recursive.
${\left\{a,b,\left(a+d\right),\left(b+d\right),\left(a+2d\right),\left(b+2d\right),\left(a+3d\right),\left(b+3d\right),\ldots \right\}}$
$d$
${\left\{a,\left(a+d\right),\left(a+2d\right),\left(a+3d\right),\left(a+4d\right)\ldots \right\}}$
$x_{1}=a,x_{2}=b;x_{i}=x_{i-2}+d$
$x_{1}=a;x_{i}=x_{i-1}+d$
* If $a=b$, then the sequence generated by using the formula $x_{1}=a,x_{2}=b;x_{i}=x_{i-2}+d$ is ${\left\{b,b,\left(b+d\right),\left(b+2d\right),\left(b+3d\right),\left(b+4d\right),\left(b+5d\right),\ldots \right\}}$ However, for a sequence to be arithmetic, each term must be added by the constant difference. Since the term in index $i=2$ is not added by $d$, the formula is not arithmetic.
${\left\{a,b,\left(a+d\right),\left(b+d\right),\left(a+2d\right),\left(b+2d\right),\left(a+3d\right),\left(b+3d\right),\ldots \right\}}$
$d$
${\left\{a,\left(a+d\right),\left(a+2d\right),\left(a+3d\right),\left(a+4d\right)\ldots \right\}}$
$x_{1}=a,x_{2}=b;x_{i}=x_{i-2}+d$
$x_{1}=a;x_{i}=x_{i-1}+d$
* If $a=b$, then the sequence generated by using the formula $x_{1}=a,x_{2}=b;x_{i}=x_{i-2}+d$ is ${\left\{b,b,\left(b+d\right),\left(b+2d\right),\left(b+3d\right),\left(b+4d\right),\left(b+5d\right),\ldots \right\}}$ However, for a sequence to be arithmetic, each term must be added by the constant difference. Since the term in index $i=2$ is not added by $d$, the formula is not arithmetic.
$x_{1}=a$
${\left\{x_{1},x_{2},x_{3},\ldots \right\}}$
${\left\{a,a,a,\ldots \right\}}$
${x_{1}=a;x_{i}=x_{i-1}+d}$
$a=a+d$
$a=a+d$
$d=0$
$d$
${a,a+1\cdot 0,a+2\cdot 0,a+3\cdot 0,a+4\cdot 0,\ldots }$
${x_{1}=a;x_{i}=x_{i-1}+d}$
$a=a+d$
$a=a+d$
$d=0$
$d$
${a,a+1\cdot 0,a+2\cdot 0,a+3\cdot 0,a+4\cdot 0,\ldots }$

#### Direct Formula

By now, you may be wondering if there is a way to find the term of an arithmetic sequence directly. Well, there is. Before giving you the formula, let us go through the motions for our general recursive arithmetic formula ${x_{1}=a;x_{k}=x_{k-1}+d}$. Let's chart the recursive arithmetic formula.
${\begin{array}{|c|c|}k&x\\\hline 1&a\\\hline 2&a+d\\\hline 3&a+2d\\\hline 4&a+3d\\\hline \vdots &\vdots \end{array}}\!$
If you think about it, the table above is basically a linear function, although starting at $k=1$ instead of $k=0$. Write out the function as $x=a+kd$. We get near our answer. Our independent variable $k$ is horizontally translated by $1$ to the right, so $x=a+(k-1)d$ is our function. In fact, we found our direct relationship. Rewrite it the way we normally write it and we found our direct formula: $x_{k}=a+(k-1)d$.
A direct formula describes how to get a term or many terms of a sequence by only using a formula that directly finds a particular value, without stating the first term or terms of a sequence.
Given an initial value $a$ and constant difference $d$, the $i$th term in an arithmetic sequence is given by the direct formula:
$x_{i}=a+(i-1)d.$
* Example 1.1.2.a: An arithmetic sequence ${\left\{22.4,29.2,36,42.8,49.6,\ldots \right\}}$ is discovered. What is the $450$th term in the sequence?
* As always, before we can determine the answer, we need to find the "rate of change" of our sequence. Since $a=22.4$, we know that $x_{i}=22.4+(i-1)d$. The second term is $29.2$, so when $i=2$, $29.2=22.4+(2-1)d$. Of course, now we can solve for $d$:
* $29.2=22.4+(2-1)d$;
* $29.2=d$;
* $6.8=d$.
* Since we now know the common difference, we can find the smallest $450$th term in the sequence. By using our direct formula for an arithmetic sequence, we can find the index in which it is possible. Since $x_{i}=22.4-6.8(i-1)$, we find the $450$th term by substituting $i=450$. Ergo, $x_{450}=22.4-6.8(450-1)$. Solve for $x_{450}$ to get the final answer.
* $x_{450}=22.4-6.8(450-1)$;
* $x_{450}=22.4-6.8(449)$;
* $x_{450}=22.4-3121.2$;
* $x_{450}=-3030.8.$
The example above would be a routine, straightforward problem in the CLEP College Algebra exam. However, as practice makes perfect, we will also have non-routine problems that involve thorough understanding of the topic and concepts and skills learned, which will make up 50% of the exam. This is why it is important to do the explorations. While they may not be on a CLEP exam, they are vital in making you think like a mathematician. The next problem will be non-routine problem.
* Example 1.1.2.b: What is the smallest index needed to find a negative term in the arithmetic sequence ${\left\{22.4,21.2,20,17.6,16.4,\ldots \right\}}$?
* As always, before we can determine the answer, we need to find the "rate of change" of our sequence. Since $a=22.4$, we know that $x_{i}=22.4+(i-1)d$. We know that the second term is $21.2$, so when $i=2$, $21.2=22.4+(2-1)d$. Of course, now we can solve for $d$:
* $21.2=22.4+(2-1)d$;
* $21.2=22.4+d$;
* $-1.2=d$.
* Since we now know the common difference, we can find the smallest term needed to reach a negative number in our sequence. By using our direct formula for an arithmetic sequence, we can find the index by solving for $i$. Since $x_{i}=22.4-1.2(i-1)$, we find the minimal index in which it is possible to have a number less than zero. Ergo, $0\geq 22.4-1.2(i-1)$. Now all we have to do is solve for $i$.
* $22.4-1.2(i-1)\leq 0$;
* $-1.2(i-1)\leq -22.4$;
* $i-1\geq {-22.4 \over -1.2}$;
* $i\geq {22.4 \over 1.2}+1$;
* $i\geq 19.{\overline {6}}$.
* Since index $i$ must be greater than $19.{\overline {6}}$, the minimal index required to find the term that is negative is at
* $i=20.$

#### CLEP Practice Problems: Check Your Understanding
### Geometric Sequence

A geometric sequence is a sequence ${\left\{x_{1},x_{2},\ldots ,x_{k},x_{k+1},\ldots \right\}}$ in which a multiplied real number $r$, called the common ratio, is multiplied to each successive term, except the first term, $x_{1}$, such that the sequence forms ${\left\{x_{1},x_{1}r,x_{1}r^{2},x_{1}r^{3},\ldots \right\}}$
As always, if you are unable to understand, try a few examples of numbers to think of in your head. Let the common ratio $r=2$ and let $x_{1}=1.5$. The next term ${x_{2}=x_{1}r}$, so $x_{2}=1.5\cdot 2=3$. If you keep the pattern going for each term of the sequence, you would get the following: ${\left\{1.5,3,6,12,24,48,\ldots \right\}}$

#### Recursive Formula

As with the arithmetic formula, you can find the recursive formula and the direct formula for a geometric sequence. Since every term is multiplied by common ratio $r$, let any term of index $i$ be represented by $x_{i}$. To find the next term requires knowing the previous term. Ergo, $x_{i}=rx_{i-1}$
Given an initial value $a$ and constant difference $d$, the $i$th term in an arithmetic sequence is given by the previous $(i-1)$th term of the sequence:
    $x_{1}=a;x_{i}=rx_{i-1}$ OR
    ${\begin{cases}x_{1}=a\\x_{i}=rx_{i-1}\end{cases}}$

#### Direct Formula

As with the arithmetic sequence, Let us chart the recursive geometric formula. ${x_{1}=a;x_{k}=rx_{k-1}}$.
${\begin{array}{|c|c|}k&x\\\hline 1&a\\\hline 2&ar\\\hline 3&ar^{2}\\\hline 4&ar^{3}\\\hline \vdots &\vdots \end{array}}\!$
If you think about it, the table above is basically an exponential function, $f(k)=ab^{k}$, starting at $k=1$. Write out the function as $x=ar^{k}$. We get near our answer. Our independent variable $k$ is translated to the right $1$ unit, so $x=ar^{k-1}$ is our function. In fact, we found our direct relationship. Rewrite it the way we normally write it and we found our direct formula: $x_{k}=ar^{k-1}$.
Given an initial value $a$ and common ratio $r$, the $i$th term in an geometric sequence is given by the direct formula:
$x_{i}=ar^{i-1}$
* Example 1.2.2.a: A geometric sequence ${\left\{22.4,11.2,5.6,2.8,1.4,\ldots \right\}}$ is discovered. What is the $18$th term in the sequence?
* To find the answer, we need to know the common ratio $r$ of the sequence above. Pick any arbitrary term in the sequence and apply it to the direct geometric formula: $x_{2}=ar^{2-1}$. Find $r$:
* $11.2=22.4r.$
* $r=0.5={1 \over 2}.$
* Knowing the value of $r$, you can find the $18$th term in the sequence by using the direct formula:
* $x_{18}=22.4\left({\frac {1}{2}}\right)^{18-1}$
* $x_{18}=22.4\left(0.00000762939\right)$
* $x_{18}=0.00017089843$
As always, these examples are things you can work through yourself or follow along so that you can see how to do a problem.
$x_{1}=a$
${a,-2a,4a,\ldots }$
$r$
$d$
$r$
$r$
$d$
$r$
$f$
$p$
$\mathbb {R}$
$p\in \mathbb {R}$
$f(x)=ax^{p}$
$p={1 \over 2}$
$a=2$
$f(x)=ax^{p}$
* multiply each value of $x$ by a common ratio $r$. Determine whether the range will also be a geometric sequence through example using any value of $r$.
* prove that multiplying the domain, $x$, by a common ratio $r$ will give a range that is geometric (where the common difference for the domain is $r_{f})$.
$f$
$r$
$x$
$r=2$
$x\cdot 2$
$f(2x)=2{\sqrt {2x}}$
$f(2x)$
${2{\sqrt {2}},\,4,\,2{\sqrt {6}},\,2{\sqrt {8}},\ldots }$
${\sqrt {2}}$
$f$
$r$
* The question is asking whether or not $f(xr)$ will give a range for $f(x)$ such that you multiply by a constant. Let's define that constant as $r_{f}$. Let $f(x)=f\left(x_{1}\right)$. Since $f\left(x_{1}\right)=a\left(x_{1}\right)^{p}$, if $x_{2}=x_{1}r$, then $f\left(x_{2}\right)=a\left(x_{2}\right)^{p}$. Since $x_{2}=x_{1}r$, function $f\left(x_{2}\right)=a\left(x_{1}r\right)^{p}$. Since any term $(ab)^{m}$ will give $a^{m}b^{m}$, function $f\left(x_{2}\right)=ax_{1}^{p}r^{p}$. Since $ax_{1}^{p}=f\left(x_{1}\right)$, function $f\left(x_{2}\right)=f\left(x_{1}\right)r^{p}$. Since $r$ and $p$ are constant, the expression $r^{p}$ will be constant. Since you are multiplying the range, $f\left(x_{1}\right)$, by a common ratio (otherwise known as a geometric sequence), given the constant $r^{p}=r_{f}$, then when multiplying the values of the domain, the range must be multiplied by a constant:
$f\left(x_{2}\right)=f\left(x_{1}\right)r_{f}.$
$f$
$r$
$x$
$r=2$
$x\cdot 2$
$f(2x)=2{\sqrt {2x}}$
$f(2x)$
${2{\sqrt {2}},\,4,\,2{\sqrt {6}},\,2{\sqrt {8}},\ldots }$
${\sqrt {2}}$
$f$
$r$
* The question is asking whether or not $f(xr)$ will give a range for $f(x)$ such that you multiply by a constant. Let's define that constant as $r_{f}$. Let $f(x)=f\left(x_{1}\right)$. Since $f\left(x_{1}\right)=a\left(x_{1}\right)^{p}$, if $x_{2}=x_{1}r$, then $f\left(x_{2}\right)=a\left(x_{2}\right)^{p}$. Since $x_{2}=x_{1}r$, function $f\left(x_{2}\right)=a\left(x_{1}r\right)^{p}$. Since any term $(ab)^{m}$ will give $a^{m}b^{m}$, function $f\left(x_{2}\right)=ax_{1}^{p}r^{p}$. Since $ax_{1}^{p}=f\left(x_{1}\right)$, function $f\left(x_{2}\right)=f\left(x_{1}\right)r^{p}$. Since $r$ and $p$ are constant, the expression $r^{p}$ will be constant. Since you are multiplying the range, $f\left(x_{1}\right)$, by a common ratio (otherwise known as a geometric sequence), given the constant $r^{p}=r_{f}$, then when multiplying the values of the domain, the range must be multiplied by a constant:
$f\left(x_{2}\right)=f\left(x_{1}\right)r_{f}.$

## Series

It is great to find these patterns to these sequences, but is that the only use we have for these sequences? As always, the answer in math is never no in regards to a low amount of utility (i.e. usefulness). A function does not only describe the pattern associated with numbers but also predicts the graph created when plotted using $x$ for its inputs and $y$ for its outputs. Using the terms of a sequence, can we determine a sum? We can, and it is called a series.
A series is the sum of the terms of a sequence.
There is always a way to find the sum of a sequence by force: add up the terms one by one and get the answer. Mathematicians are lazy people and don't want to do more work than necessary. That is to say, mathematicians work smart, not hard.

### Arithmetic Series

* Example 2.1.1.a: What is ${1+2+3+\ldots +97+98+99+100}$?
* This is a classic example among many mathematicians. You may be reaching for your calculator, but that will definitely take a long time to punch in all those numbers, and you may accidently make mistakes while writing out the expression. The question is, how do we solve this in the short amount of time we are given on the CLEP exam? Notice that there is a pattern at play: the first term of the series, $1$, plus the last term of the series, $100$, will give us $101$. The second term of the series, $2$, plus the second-to-last term of the series, $99$, will give us $101$. In fact, for all $100$ terms in the series, each pair of numbers, according to their "placement," adds to $101$. Since there are $50$ pairs in this series, and the sum of each pair of numbers always yields $101$,
* $101\cdot 50=5050{\text{ is the sum of the expression above.}}$
Let us make a conjecture for the above statement. After all, we are not robots that are simply fed a method and do something the same way. We will contract parts of the above expression in the question above to see if our method will be useful for any number of terms. Let $n$ be the number of terms in the arithmetic series $A_{E}$, let $S$ be the sum of $A_{E}$, and let $S_{m}$ be the column in which the same method we used in Example 2.1.1.a will, hopefully, get us the sum $S$.
${\begin{array}{|c|c|}n&A_{E}&S&S_{m}\\\hline 2&1+2&3&(1+2)\cdot 1=3\\\hline 3&1+2+3&6&(1+3)\cdot 1.5=6\\\hline 4&1+2+3+4&10&(1+4)\cdot 2=10\\\hline 5&1+2+3+4+5&15&(1+5)\cdot 2.5=15\\\hline \vdots &\vdots &\vdots &\vdots \\\hline n&1+2+\ldots +(n-1)+n&S&(1+n)\cdot {n \over 2}=S\end{array}}\!$
Notice how the odd terms have this weird (or perhaps odd) behavior where not all pairs of terms have the same sum (because there is not an even amount of pairs). While it may seem that using the method in Example 2.1.1.a will work for all odd number of terms based on the table above, maybe it is not true for some really high odd term. Therefore, it is important to prove this is true. For now, let us simply define the formula to be true for both even and odd number of terms.
For any finite sequence ${\left\{x_{i}\right\}}$, where $x_{i}=a+(i-1)d$, that contains $k$ terms, the sum of each term is
$S_{A}={a+x_{k} \over 2}\cdot k.$
A finite set of things is something that has a determined number of anythng within that set. An infinite set is a set not described by a determined number of the amount within that set.
Note: the above formula is not known as the "Gaussian method" among mathematicians. This wikibooks will simply refer to this method as the "Gaussian method."$^{[seefootnote1.]}$
* Example 2.1.1.b: Knowing the sum to an arithmetic series is $600$ and the first term is $50$ but the last term is $250$, what is the constant difference?
* This seems like an impossible problem to someone who does not know about the Gaussian method or the properties of an arithmetic sequence. However, since you have paid attention, you can figure out for yourself. For the purposes of giving an example, this Wikibooks will explain. We do not know how many terms there are in the sequence, so let's find out how many terms there are. Since
* $S_{A}={a+x_{k} \over 2}\cdot k,$
* $600={50+250 \over 2}\cdot k.$
* This subject is not called College Algebra for nothing, so do some algebra.
* $600={300 \over 2}\cdot k$
* $600=150k$
* $40=k$
* Note, however, that we are not looking for the number of terms in the series. We want to know the constant difference of the arithmetic series. Therefore, use the arithmetic sequence direct formula: $x_{k}=a+d(k-1)$!
* $x_{40}=250$
* $250=50+d(40-1)$
* $200=39d$
* ${200 \over 39}=d$
* $5{5 \over 39}=d$
Example 2.2.1.b helps us know how to find the number of terms and the constant difference. Sometimes, you may simply not know one or both of the information you needed to find in Example 2.2.1.b., more often the number of terms. The next example helps illustrate the usefulness of knowing about the properties of arithmetic series and arithmetic sequences.
* Example 2.1.1.c: What is the sum of the arithmetic series ${4+9+14+\ldots +2,494+2,499+2,504}$?
* To find the sum, we need to know how many terms there are; otherwise we cannot use the Gaussian method. Because we do not know the sum of the series, let's use the direct arithmetic sequence formula (since we know which terms correspond to place). We want to find the number of terms, $k$, so use the last term ($x_{k}=2,504$).
* First, find the constant difference. To get from $4$ to $9$, we need to add $5$ to $4$. Therefore, the constant difference is $d=5$. To put it more formally: if recursive formula $x_{1}=4;x_{2}=4+d$ and $x_{2}=9$, then $x_{1}=4;9=4+d$ means $d=5$.
* Second, find the number of terms. If $x_{k}=2,504$, $d=5$, and $x_{1}=a=4$, then $2,504=4+5(k-1)$. Solve for $k$:
* $2,500=5(k-1)$
* $500=k-1$
* $501=k.$
* Finally, use the Gaussian Method:
* $S_{A}={4+2,504 \over 2}\cdot 501$
* $S_{A}={2,508 \over 2}\cdot 501$
* $S_{A}=1,254\cdot 501$
* $S_{A}=628,254.$
As you can see, an arithmetic series can be useful in describing any type of sequence. Now that we have sufficiently explored arithmetic series, let's prove that the Gaussian method works for odd number of terms.
Note: what you are about to learn is NOT required for the curriculum. If you do not understand this proof, do not worry, for it does not matter for the CLEP exam. These proofs are only to build a mathematical understanding of the concepts. As such, you may skip these if you want.
Given there are $k$ terms in the arithmetic sequence, $x_{k}=a+d(k-1)$.
Adding one more term makes the sequence even because $k$ is odd: $x_{k+1}=a+d([k+1]-1)=a+dk$.
Since $x_{k+1}=a+dk$, adding $x_{k+1}$ to the series makes that the last term, so ${a+a+dk \over 2}\cdot (k+1)$.
Because ${x \over z}=xz^{-1}$ and $xz=zx$, $2^{-1}\cdot (a+a+dk)\cdot (k+1)=(a+a+dk)\cdot 2^{-1}(k+1)=(a+a+dk)\cdot {k-1 \over 2}$.
$-(a+dk)+(a+a+dk)\cdot {k-1 \over 2}$ will help us find the number of terms we are looking for:
    ${{(2a+dk)(k-1) \over 2}-a-dk={2ak+2a+dk+dk^{2} \over 2}-a-dk={2ak+2a+dk+dk^{2} \over 2}-{2a-2dk \over 2}=}$
    ${{2ak+2a+dk+dk^{2}-2a-2dk \over 2}={2ak-dk+dk^{2} \over 2}={k(2a-d+dk) \over 2}={k \over 2}\cdot (2a-d+dk).}$
    Since $-d+dk$ has multiplier $d$ for both terms, $d(k-1)$.
    ${[2a+d(k-1)]\cdot {k \over 2}=[a+a+d(k-1)]\cdot {k \over 2}}.$
    Because $x_{k}=a+d(k-1)$, ${[a+a+d(k-1)]\cdot {k \over 2}=\left(a+x_{k}\right)\cdot {k \over 2}={a+x_{k} \over 2}\cdot k}$.
${a+x_{k} \over 2}\cdot k$ is the same formula we started with; ergo, the same formula works for odd number terms and even number terms.

#### CLEP Practice Problems: Check Your Understanding

Note, this practice quiz is unfinished.

#### Proof of Arithmetic Series

Note: what you are about to learn is NOT required for the curriculum. If you do not understand this proof, do not worry, for it does not matter for the CLEP exam. These proofs are only to build a mathematical understanding of the concepts. As such, you may skip these if you want.

### Geometric Series

While it is amazing one can find the sum of an arithmetic sequence, what about the sum of a geometric sequence? Thankfully, mathematicians have found a way to calculate this idea. This theorem (not conjecture, as you will hopefully see with the geometric series proof) is one of many ways mathematicians have found solutions to problems coming from the abstract. As always, it is best to understand the way you can use a formula first before getting the formula introduced. To invoke deeper learning, we want to understand the concept, not reciprocate the formula outloud on a test and say, "I know the answer."
* Example 2.2.1.a: What is the sum of the geometric series ${1+2+4+8+\ldots +512}$? Let's define a sum $S$ for the geometric sequence: ${1+2+4+8+\ldots +512=S}$
* Notice that all the terms except $x_{1}=1$ has a factor $2$. Therefore, move the first term to the other side by subtracting $1$: ${2+4+\ldots +512=S-1}$. Then, factor the two on the left side: ${2(1+2+4+\ldots +256)=S-1}$
* The key discovery of this exercise is noticing that ${1+2+4+\ldots +256}$ is $S$ but without $512$, so given that ${S=1+2+4+8+\ldots +512}$,
* ${1+2+4+\ldots +256}=S-512$
* Solve using elementary algebra:
* ${2S-1024=S-1}$
* ${2S-S=-1+1024}$
* ${S=1023}$
Notice how we solved the problem. We said that the sum must equal something positive, so we determined that if it does equal something, we could figure it out in some way without having to do any long calculations. Can we use this method for some general geometric sequence? This how we determine a formula. In your Exploration, you will be tasked with presenting a proof for the formula of the geometric series. We will still give you the formula, but you must prove it yourself in the next exploration. Of course, we will also present another way to prove the geometric series formula in the next section.
$a$
$r$
$k$
$S={a\left(1-r^{k}\right) \over 1-r}$
$k$
$\left\{a_{i}\right\}=ar^{i-1}$
$k$
$S=a+ar+ar^{2}+\ldots +ar^{k-1}$
We know that $S-a$ will make it so that all terms to the right of the equal sign have factor $r$. Ergo,
$S-a=r\left(a+ar+ar^{2}+\ldots +ar^{k-2}\right).$
Note that $S-ar^{k-1}=a+ar+ar^{2}+\ldots +ar^{k-2}$, so divide by $r$ and apply the identity to the expression to yield
${S-a \over r}=S-ar^{k-1}.$
Subtract $S$ and write the expression to the left as a fraction:
${S-a \over r}-{S \over 1}={S-a-Sr \over r}=-ar^{k-1}.$
Multiply by $r$ and add $a$:
$S-Sr=a-ar^{k}.$
Factor $S$ on the left, factor $a$ on the right and divide by $1-r$:
$S={a\left(1-r^{k}\right) \over 1-r}.$
$k$
$\left\{a_{i}\right\}=ar^{i-1}$
$k$
$S=a+ar+ar^{2}+\ldots +ar^{k-1}$
We know that $S-a$ will make it so that all terms to the right of the equal sign have factor $r$. Ergo,
$S-a=r\left(a+ar+ar^{2}+\ldots +ar^{k-2}\right).$
Note that $S-ar^{k-1}=a+ar+ar^{2}+\ldots +ar^{k-2}$, so divide by $r$ and apply the identity to the expression to yield
${S-a \over r}=S-ar^{k-1}.$
Subtract $S$ and write the expression to the left as a fraction:
${S-a \over r}-{S \over 1}={S-a-Sr \over r}=-ar^{k-1}.$
Multiply by $r$ and add $a$:
$S-Sr=a-ar^{k}.$
Factor $S$ on the left, factor $a$ on the right and divide by $1-r$:
$S={a\left(1-r^{k}\right) \over 1-r}.$
For any finite geometric sequence ${\left\{x_{i}\right\}}$, where $x_{i}=ar^{i-1}$, that contains $k$ terms, the product of each term is
$S_{G}={a\left(r^{k}-1\right) \over r-1}$
for this Wikibooks$^{[seefootnote2]}$.

#### Proof of Geometric Series

Note: what you are about to learn is NOT required for the curriculum. If you do not understand these concepts, do not worry, for it does not matter for the CLEP exam. These proofs are only to build a mathematical understanding of these concepts. As such, you may skip these if you want.
By definition, a geometric series has the sequence $x_{i}=x_{1}\cdot r^{i-1}$, with $x_{1}=a$. As a result, the following is true: $a+ar+ar^{2}+\ldots +ar^{k-2}+ar^{k-1}$.
Let $a+ar+ar^{2}+\ldots +ar^{k-2}+ar^{k-1}=S$.
Add $ar^{k}$ to both sides: $a+ar+ar^{2}+\ldots +ar^{k-2}+ar^{k-1}+ar^{k}=S+ar^{k}.$
Subtract $a$ to the other side: $ar+ar^{2}+\ldots +ar^{k-2}+ar^{k-1}+ar^{k}=S+ar^{k}-a.$
Factor the common ratio $r$ to all terms in the left side of the equation: $r\left(a+ar+ar^{2}+\ldots +ar^{k-2}+ar^{k-1}\right)=S+ar^{k}-a.$
Factor $a$ for the expression $ar^{k}-a$ to the right side of the equation: $r\left(a+ar+ar^{2}+\ldots +ar^{k-2}+ar^{k-1}\right)=S+a\left(r^{k}-1\right).$
Notice that because $a+ar+ar^{2}+\ldots +ar^{k-2}+ar^{k-1}=S$, we can substitute that into the expression on the left side: $rS=S+a\left(r^{k}-1\right).$
Finally, solve for $S$: $rS-S=a\left(r^{k}-1\right)$
$S(r-1)=a\left(r^{k}-1\right)$
$S={a\left(r^{k}-1\right) \over r-1}.$
Therefore, the sum of a geometric series must be $S={a\left(r^{k}-1\right) \over r-1}$.

### Sigma Notation

How many of you are tired of writing out the expressions? How many of you are tired of figuring out how many expressions there are in a series? Where is this short-hand that we conveniently have for an arithmetic or geometric sequence? Thankfully, all of your answers are coming soon.
Say you want to write out the expression $\underbrace {1+1+1+1+\ldots +1+1} _{500}$, in which there are $500$ terms in the series, each of which are the number $1$. We hopefully can evaluate the series without having to think too much about it, or even think about a short-hand about this expression immediately without too much thought. However, the purpose of this easy exercise is to introduce a new notation.
Let $\sum$ denote the sum of a series. Located at the bottom of the symbol is the start of the series at index $i=a$ (blue) and the top of the symbol is the last index of the series, $k$ (red). The term inside the parentheses represents the sequence that the series follows, $f(i)$ (orange). The use of the notation of is shown below $\sum _{\color {Blue}i=a}^{\color {Red}k}\left({\color {Orange}f\left(i\right)}\right).$Here, we know that $\left\{1,1,1,1,\ldots 1,1\right\}$ must have the sequence $x_{i}=1+0(i-1)=1=f\left(i\right)$.
Interpreting the meaning of $f(i)$ tells us that for any number in the sequence, the $i$th term of the expression is $1$. Knowing that the sequence starts at index $1$, meaning $i=1$, and there are $500$ terms, meaning $k=500$, we may write the sigma notation as below: $\sum _{\color {Blue}i=1}^{\color {Red}500}\left({\color {Orange}1}\right).$
It is important to write the sigma representation having $f(i)$ in parenthesis. Often, if you don't have it, you can confuse other terms by accident. It is for this reason that this Wikibooks recommends writing the $f(i)$ term in parentheses.
For any series that has formula $f(i)$, starting from arbitrary point $i=a$, and given $k$ terms, the sigma $\left(\sum \right)$ representation is written as follows: $\sum _{i=a}^{k}\left(f\left(i\right)\right).$
Some more ways in which sigma notation is written is often a short-hand, especially when handwriting. This should be used if you want to save time.
* $\sum _{i=a}^{k}\left(f(i)\right)=\sum _{a\leq i\leq k}\left(f(i)\right)$
* $\sum _{i=a}^{\infty }\left(f(i)\right)=\sum _{i\geq a}\left(f(i)\right)$
Before we jump into our exploration, it may be a good idea to introduce some more expressions that can be rewritten into its sigma form equivalents. Some of the expressions may not even follow from a formula we have seen so far. Nevertheless, let us continue with this concept.
* Example 2.3.1.a: Write the sequence into its sigma form equivalent: ${\frac {2}{3}}+{\frac {3}{4}}+{\frac {4}{5}}+\cdots +{\frac {70}{71}}+{\frac {71}{72}}$
* A good idea before getting started is to look for any possible patterns. Because you are working with sigma notation, check whether there has to be a pattern involving addition.
* Look at the numerator of the first two terms. Let us assume we are starting at index $i=1$. To get from the first term in the numerator to the second term in the numerator, one simply has to add 1 to the first term. Similarly, to get from the first term in the denominator to the second term in the denominator, one simply has to add 2 to the first term. Both of these are true on a term-by-term basis.
* $f(i)={\frac {i+1}{i+2}}$. This is easy to verify for every term in its respective index.
* Because this function is true, let us use this to find the number of terms in the expression (i.e., look at the final term in the expression).
* Let $f(k)={\frac {71}{72}}={\frac {k+1}{k+2}}$. There are many ways to solve for $k$, the last term of the expression, in this equation. We will show the standard way.
* ${\frac {k+1}{k+2}}={\frac {71}{72}}$
* $\Leftrightarrow 72\cdot (k+1)=71\cdot (k+2)$
* $\Leftrightarrow 72k+72=71k+142$
* $\Leftrightarrow k=70$
* We have demonstrated this expression involves 70 terms. With the index, number of terms, and function ready, we may now write the sigma equivalent:
* $\sum _{i=1}^{70}\left({\frac {i+1}{i+2}}\right)\blacksquare$
* Alternatively, one can change the index to get a completely different sigma notation. Notice how when the index starts at an $i>1$, the formula and the final term's index also changes. Here, we will look specifically at $i=2$. Keeping in mind the index represents the placement along the series, going from 2 to 3 is simply a matter of changing along with the numerator. Because the denominator is simply the same idea except displaced by 1 more than the index, the denominator follows a pattern of $i+1$. Therefore, the function of the sigma notation is
* $g(i)={\frac {i}{i+1}}$. This is easy to verify for every term in its respective index.
* Use this function to find the the final term's index.
* Let $f(k)={\frac {71}{72}}={\frac {k}{k+1}}$. We will solve this one the simplest way. Because both the numerator and the denominator are constant, non-zero terms, and $k$ is simply an expression of the index for its final term, one can simply set the numerator equal to the other numerator, and the same is true for the denominator. That is,
* ${\begin{cases}k=71\\k+1=72\end{cases}}$
* This is perhaps the easiest systems of equations one had to deal with this year learning college algebra. From this, it is obvious $k=71$.
* Knowing the index starting point, function, and final index, the sigma form of this situation is
* $\sum _{i=2}^{71}\left({\frac {i}{i+1}}\right)\blacksquare$
* Notice how the pattern changed. Quite intriguing, would you not say?
Some expressions are too hard to determine simply from looking. These are likely going to involve factors multiplied to another. THe best way to determine such a pattern would be to divide terms to see what the factors are.
* Example 2.3.1.b: Write the sequence into its sigma form equivalent: $0-2-2+0+4+10+\ldots$
* There seems to be no pattern here. The best thing to do is to determine any similarities between terms and see if there seems to be a pattern from there.
* Notice how the first and fourth term are zero. This means that there is a term that allows us to obtain a value of zero. Let us assume they are linear functions. If $k=0$ is the index of the first term and $k=3$ is the third term, then a possible summation function is $\sum _{k\geq 0}\left(k(k-3)\right)$. Let us apply this and see if it works.
* $k=0\Rightarrow 0$
* $k=1\Rightarrow -2$
* $k=2\Rightarrow -2$
* $k=3\Rightarrow 0$
* $k=4\Rightarrow 4$
* $k=5\Rightarrow 10$
* It seems this function works for each term. Therefore, this works.
* $\sum _{k=0}^{\infty }\left(k(k-3)\right)=\sum _{k=0}^{\infty }\left(k^{2}-3k\right)\blacksquare$
More examples will be added later. The next exploration will ask you write out a variety of expressions using sigma form:
(a) $1+3+5+7+9+\ldots +541+543$ (b) $1+{1 \over 2}+{1 \over 4}+{1 \over 8}$ (c) $1+2+3+\ldots +99+100+102+104+\ldots +200$
$100-20+4-{4 \over 5}+{4 \over 25}-100-120-140$
(a) Remember how we needed to find out how many terms there are in a sequence and the constant difference to find the sum of an arithmetic sequence? It is the same here. The sigma representation, well, represents the adding of each term in the series. In which case, we need to know the constant difference and the number of terms in the series to determine any validity from it. Note how $\left\{1,\,3,\,5,\,\ldots ,\,543\right\}$ is an arithmetic sequence that has the common difference two. Therefore, given the first term is $1$, the arithmetic sequence formula is $x_{i}=1+2(i-1)=f(i)$. From this, we can determine the final term's position: $543=1+2(i-1)$. After solving for $i$, you find that there are $i=273$ terms in the sequence. As such, you may finally write out the final answer: $\sum _{i=1}^{273}\left(1+2\left(i-1\right)\right).$ Note, because you can start counting the terms at $i=0$, you may rewrite the expression as follows: $\sum _{i=0}^{272}\left(1+2i\right).$
(b) Use the same idea from item (a) to rewrite the expression into its sigma equivalent. Notice that the series shows a geometric series, with common ratio of ${1 \over 2}$ starting at $1$. Therefore, $f(i)=\left({1 \over 2}\right)^{i-1}$. Since there is no need to use a formula to find out how many terms there are in the series, simply count and you have your answer: $\sum _{i=1}^{4}\left({1 \over 2}\right)^{i-1}.$ Again, you may rewrite the sigma form equivalent starting at $i=0$: $\sum _{i=0}^{3}\left({1 \over 2}\right)^{i}.$
(c) There are two series shown in the expression. Separate each series within as the following: $f_{1}\left(i\right)=1+2+3+\ldots +99$ and $f_{2}\left(i\right)=100+102+104+\ldots +200$. Both $f_{1}$ and $f_{2}$ are arithmetic sequences but with different differences and starting numbers. Instead of explaining each one, only the formula will be provided for each sequence: $f_{1}(i)=1+1(i-1)$ and $f_{2}(i)=100+2(i-1)$. Note that the starting number and end numbers are not appropriate to use in the same sigma notation, so separate them into two sigma representations and add them together. Thus, the sigma representation for this expression is $\sum _{i=1}^{99}(1+1(i-1))+\sum _{i=1}^{51}(100+2(i-1)).$ Do you see why we ask you to write the formula using parentheses? Also, you may have separated the series differently. In which case, if $f_{1}\left(i\right)=1+2+3+\ldots +99+100$ and $f_{2}\left(i\right)=102+104+\ldots +200$, then $f_{1}(i)=1+1(i-1)$ and $f_{2}(i)=102+2(i-1)$, where $f_{1}$ has $100$ terms and $f_{1}$ has $50$ terms. Therefore, another representation is also appropriate: $\sum _{i=1}^{100}(1+1(i-1))+\sum _{i=1}^{50}(100+2(i-1)).$
(d) There are two series shown in the expression. Separate each series within as the following: $f_{1}\left(i\right)=100-20+4-{4 \over 5}+{4 \over 25}$ and $f_{2}\left(i\right)=-100-120-140$. While $f_{1}$ is geometric, with common ratio $-{1 \over 5}$, $f_{2}$ is arithmetic, with constant difference $-20$. Instead of explaining each one, only the formula will be provided for each sequence: $f_{1}(i)=100\left(-{1 \over 5}\right)^{i-1}$ and $f_{2}(i)=-100-20(i-1)$. Because of the difference in formula and difference in starting numbers used, it is impossible to put them together in the same sum. Therefore, the sigma form of the series is $\sum _{i=1}^{5}\left(100\left(-{1 \over 5}\right)^{i-1}\right)+\sum _{i=1}^{3}(-100-20(i-1)).$ Notice how you added the other term instead of substracting. If you were to subtract the second sigma representation you would in effect flip the sign on every other term. As such, the second series representation would be wrong unless you were to change it to this:
$\sum _{i=1}^{5}\left(100\left(-{1 \over 5}\right)^{i-1}\right)-\sum _{i=1}^{3}(100+20(i-1)).$
(a) Remember how we needed to find out how many terms there are in a sequence and the constant difference to find the sum of an arithmetic sequence? It is the same here. The sigma representation, well, represents the adding of each term in the series. In which case, we need to know the constant difference and the number of terms in the series to determine any validity from it. Note how $\left\{1,\,3,\,5,\,\ldots ,\,543\right\}$ is an arithmetic sequence that has the common difference two. Therefore, given the first term is $1$, the arithmetic sequence formula is $x_{i}=1+2(i-1)=f(i)$. From this, we can determine the final term's position: $543=1+2(i-1)$. After solving for $i$, you find that there are $i=273$ terms in the sequence. As such, you may finally write out the final answer: $\sum _{i=1}^{273}\left(1+2\left(i-1\right)\right).$ Note, because you can start counting the terms at $i=0$, you may rewrite the expression as follows: $\sum _{i=0}^{272}\left(1+2i\right).$
(b) Use the same idea from item (a) to rewrite the expression into its sigma equivalent. Notice that the series shows a geometric series, with common ratio of ${1 \over 2}$ starting at $1$. Therefore, $f(i)=\left({1 \over 2}\right)^{i-1}$. Since there is no need to use a formula to find out how many terms there are in the series, simply count and you have your answer: $\sum _{i=1}^{4}\left({1 \over 2}\right)^{i-1}.$ Again, you may rewrite the sigma form equivalent starting at $i=0$: $\sum _{i=0}^{3}\left({1 \over 2}\right)^{i}.$
(c) There are two series shown in the expression. Separate each series within as the following: $f_{1}\left(i\right)=1+2+3+\ldots +99$ and $f_{2}\left(i\right)=100+102+104+\ldots +200$. Both $f_{1}$ and $f_{2}$ are arithmetic sequences but with different differences and starting numbers. Instead of explaining each one, only the formula will be provided for each sequence: $f_{1}(i)=1+1(i-1)$ and $f_{2}(i)=100+2(i-1)$. Note that the starting number and end numbers are not appropriate to use in the same sigma notation, so separate them into two sigma representations and add them together. Thus, the sigma representation for this expression is $\sum _{i=1}^{99}(1+1(i-1))+\sum _{i=1}^{51}(100+2(i-1)).$ Do you see why we ask you to write the formula using parentheses? Also, you may have separated the series differently. In which case, if $f_{1}\left(i\right)=1+2+3+\ldots +99+100$ and $f_{2}\left(i\right)=102+104+\ldots +200$, then $f_{1}(i)=1+1(i-1)$ and $f_{2}(i)=102+2(i-1)$, where $f_{1}$ has $100$ terms and $f_{1}$ has $50$ terms. Therefore, another representation is also appropriate: $\sum _{i=1}^{100}(1+1(i-1))+\sum _{i=1}^{50}(100+2(i-1)).$
(d) There are two series shown in the expression. Separate each series within as the following: $f_{1}\left(i\right)=100-20+4-{4 \over 5}+{4 \over 25}$ and $f_{2}\left(i\right)=-100-120-140$. While $f_{1}$ is geometric, with common ratio $-{1 \over 5}$, $f_{2}$ is arithmetic, with constant difference $-20$. Instead of explaining each one, only the formula will be provided for each sequence: $f_{1}(i)=100\left(-{1 \over 5}\right)^{i-1}$ and $f_{2}(i)=-100-20(i-1)$. Because of the difference in formula and difference in starting numbers used, it is impossible to put them together in the same sum. Therefore, the sigma form of the series is $\sum _{i=1}^{5}\left(100\left(-{1 \over 5}\right)^{i-1}\right)+\sum _{i=1}^{3}(-100-20(i-1)).$ Notice how you added the other term instead of substracting. If you were to subtract the second sigma representation you would in effect flip the sign on every other term. As such, the second series representation would be wrong unless you were to change it to this:
$\sum _{i=1}^{5}\left(100\left(-{1 \over 5}\right)^{i-1}\right)-\sum _{i=1}^{3}(100+20(i-1)).$

#### Sigma Simplification Techniques

In some textbooks, this section would be called rules. We call them what they really are: techniques to simplify common sigma representations.
For any series that has constant term $c$, starting from index $i=1$, and given $k$ terms,
$\sum _{i=1}^{k}\left(c\right)=ck.$
As always, if you are confused, write out some terms. The above sigma representation states that for any constant term $c$ added $k$ times, the resulting sum is equivalent to $\underbrace {c+c+c+\ldots +c+c+c} _{k}=ck$. Never forget that multiplication is repeated addition. This axiom you heard in elementary school is still important to this day.
For any series that has constant term $c$ multiplied by changing index $i$, starting from index $i=1$, and given $k$ total terms,
$\sum _{i=1}^{k}\left(ci\right)=c\sum _{i=1}^{k}\left(i\right).$
This second "rule" can be simplified as the following. Since $\sum _{i=1}^{k}\left(ci\right)$, we can conclude that $\sum _{i=1}^{k}\left(ci\right)=1c+2c+3c+\ldots +c(k-2)+c(k-1)+ck$. Each term in the expression $1c+2c+3c+\ldots +c(k-2)+c(k-1)+ck$ has common factor $c$, so $c(1+2+3+\ldots +(k-2)+(k-1)+k)$. Notice that each term on the inside of the parenthesis is a basic series where $i$ starting at $1$ goes to final term $k$ as a sum, so
$\sum _{i=1}^{k}\left(ci\right)=c\sum _{i=1}^{k}\left(i\right).$
For any series that has constant term $c$ in addition to changing index $i$, starting from index $i=1$, and given $k$ total terms,
$\sum _{i=1}^{k}\left(c+i\right)=kc+\sum _{i=1}^{k}\left(i\right).$
This one is harder to see why it is true. As always, however, try it out by hand. Remember that sigma notations are short hands of sums that have some sort of formula, so write out a term in case you are ever confused. Notice that $\sum _{i=1}^{k}\left(c+i\right)=(c+1)+(c+2)+(c+3)+(c+4)+\ldots +(c+k-1)+(c+k)$. Because adding is commutative, and because the parentheses do not change the sum of the series, you may group terms such that $(c+1)+(c+2)+(c+3)+(c+4)+\ldots +(c+k-1)+(c+k)=\underbrace {c+c+c+c+\ldots +c+c} _{k}+(1+2+3+\ldots +k-1+k)$ We have already determined the sigma notation for each grouping, so we may put it together to say the following is true:
$\sum _{i=1}^{k}\left(c+i\right)=kc+\sum _{i=1}^{k}\left(i\right).$
For any series that has constant term $c$ in addition to the range of the function of $i$ or $f\left(i\right)$, starting from index $i=1$, and given $k$ total terms,
$\sum _{i=1}^{k}\left(c+f\left(i\right)\right)=kc+\sum _{i=1}^{k}\left(f\left(i\right)\right).$
Realize that the above "rule" is simply an extension of the previous one. If $f(i)=i$, then $\sum _{i=1}^{k}\left(c+f\left(i\right)\right)=kc+\sum _{i=1}^{k}\left(f\left(i\right)\right)$. Realize that any function can comply with this rule. By extension:
$\sum _{i=1}^{k}\left(cf\left(i\right)\right)=c\sum _{i=1}^{k}\left(f\left(i\right)\right)$
Using this general knowledge, one can also argue the next following "rule":
For any series that has constant term $a$ multiplied by the function of $i$ or $f\left(i\right)$, starting from index $i=1$, when adding constant $c$ to the product, and given $k$ total terms,
$\sum _{i=1}^{k}\left(c+af\left(i\right)\right)=kc+a\sum _{i=1}^{k}\left(f\left(i\right)\right).$
Given the way we found the same identities, you may do the proof for this one by yourself as an exercise in sigma notation. The best way to understand this new notation is to practice it. (Of course, this Wikibooks will provide plenty of practice.)
Along with that practice, you may also try to show that the identity for the next "rule" is also true:
For any series in which the formula is two different arbitrary functions: $f\left(i\right)$ and $g\left(i\right)$, starting from index $i=1$, and given $k$ total terms,
$\sum _{i=1}^{k}\left(f\left(i\right)+g\left(i\right)\right)=\sum _{i=1}^{k}\left(f\left(i\right)\right)+\sum _{i=1}^{k}\left(g\left(i\right)\right).$
With this, you now have a sufficient foundation for the necessary tools needed to prove and disprove statements as well as create your own identity. Before we move on the next section, we must mention the sigma identity for each of the series type we learned in this wikibooks.

#### Sigma Identities for Series Types

What is the sigma identity of an arithmetic series? Before we give you the identity, it is important to understand arithmetic series. If the person reading this wikibooks has jumped from one section to the other, we may recommend you read anything from the previous if the current reasoning seems to not make sense. First, how do we write an arithmetic sequence using a formula? Like this: $x_{k}=a+(k-1)d.$ The formula for how to find an arithmetic series is this: $S_{a}={a+x_{k} \over 2}k.$ Notice how $x_{k}$ is in both formulas, so substitute $x_{k}$ for $a+(k-1)d$ and you get: $S_{a}={2a+(k-1)d \over 2}k=ak+{(k-1)d \over 2}k=\left[2a+(k-1)d\right]{k \over 2}.$ Because $x_{k}=a+(k-1)d.$ is also a formula that helps us find each term in the series, we have finally created our sigma identity:
An arithmetic series in which the formula is $f(i)=a+(i-1)d$, starting from index $i=1$, and given $k$ total terms,
$\sum _{i=1}^{k}\left(a+(i-1)d\right)={2a+(k-1)d \over 2}\cdot k.$
We can use the same process to find the sum of a geometric series. You may consider this your next exercise proving that the sigma form of the geometric series is the same one shown below.
A geometric series in which the formula is $f(i)=ar^{i-1}$, starting from index $i=1$, and given $k$ total terms,
$\sum _{i=1}^{k}\left(ar^{i-1}\right)={a\left(r^{k}-1\right) \over r-1}.$

#### Check Your Understanding

The problems below are basically comprehension questions. If you can do all the problems below, you understand all you need to know about sigma notations. Keep in mind, these problems are much harder than usual CLEP problems; however, being able to do these problems with proficiency proves understanding of the material at a deeper level, which means you are better prepared for easier problems and harder problems down the line.
1. Find the smallest $x$ necessary to give the smallest real number for the expression below. After, simplify and calculate for the smallest real value. Do not round your answer, and do not use a calculator or program.
$\log _{5}\left(\sum _{i=1}^{x}\left(-500+{1 \over 3}i\right)\right)$
$\log _{a}c=b$
$a^{b}=c$
$\left\{a\vert a>0,\,a\neq 1\right\},\,\left\{b\vert b>0,\,b\in \mathbb {R} \right\}{\text{ and}},\,c>0$
$c>0$
$\sum _{i=1}^{x}\left(-500+{1 \over 3}i\right)>0$ Separate the two expressions inside the sum by applying a property rule, given constant $C=-500$ and $f(i)={1 \over 3}i$: $\sum _{i=1}^{x}\left(-500+{1 \over 3}i\right)=-500x+\sum _{i=1}^{x}{1 \over 3}i>0$ Since $\sum _{k=1}^{n}Cf(i)=C\sum _{k=1}^{n}f(i)$, and $f(i)$ is arithmetic, it must be that case that $\sum _{i=1}^{x}{1 \over 3}i={1 \over 3}\cdot {1+x \over 2}x={1 \over 3}\cdot {x(x+1) \over 2}={x(x+1) \over 6}.$ From there, put the sum of the two expressions over the same denominator and simplify:
    $-500x+{x(x+1) \over 6}>0$
    $\Rightarrow {-3000x \over 6}+{x(x+1) \over 6}>0$
    $\Rightarrow {-3000x+x^{2}+x \over 6}>0$
    $\Rightarrow {-2999x+x^{2} \over 6}>0$
    $\Rightarrow -2999x+x^{2}>0$
    $\Rightarrow x(x-2999)>0$
Finally, note that the zero factor property gives $ab=0$ if and only if $a=0$ or $b=0$ or both. However, for a zero factor property in which the two multipliers give a positive product, it must be the case that either $a$ and $b$ are positive or $a$ and $b$ are negative. Through that logic, we can determine four possible solutions: $x>0$ and $x>2999$, OR $x<0$ and $x<2999$. Despite the four solutions, only one solution must be true: $x>2999$ because sigma notations only work through $x\in \mathbb {Z^{+}}$ or $x$ being the set of positive integers in the situation described in the problem. As such, all $x$ greater than $2999$ gives an acceptable solution. However, this is only the first part of the problem. Now comes the easy part.
The smallest number must be $x=3000$ in the situation described above:
    ${\begin{aligned}\log _{5}\left(\sum _{i=1}^{3,000}\left(-500+{1 \over 3}i\right)\right)&=\log _{5}\left(-500(3,000)+{3,000(3,000+1) \over 6}\right)\\&=\log _{5}\left(-1,500,000+500(3000+1)\right)\\&=\log _{5}\left(-1,500,000+1,500,000+500)\right)\\&=\log _{5}\left(500\right)\end{aligned}}$
Finally, simplify $\log _{5}\left(500\right)$ to get the final answer:
    ${\begin{aligned}\log _{5}\left(500\right)&=\log _{5}\left(100\cdot 5\right)\\&=\log _{5}\left(10^{2}\cdot 5\right)\\&=\log _{5}\left((5\cdot 2)^{2}\cdot 5\right)\\&=\log _{5}\left(5^{2}\cdot 2^{2}\cdot 5\right)\\&=\log _{5}\left(2^{2}\cdot 5^{3}\right)\\&=3\log _{5}\left(5\right)+2\log _{5}\left(2\right)+1\\&=3+2\log _{5}\left(2\right)\blacksquare \end{aligned}}$
$\log _{a}c=b$
$a^{b}=c$
$\left\{a\vert a>0,\,a\neq 1\right\},\,\left\{b\vert b>0,\,b\in \mathbb {R} \right\}{\text{ and}},\,c>0$
$c>0$
$\sum _{i=1}^{x}\left(-500+{1 \over 3}i\right)>0$ Separate the two expressions inside the sum by applying a property rule, given constant $C=-500$ and $f(i)={1 \over 3}i$: $\sum _{i=1}^{x}\left(-500+{1 \over 3}i\right)=-500x+\sum _{i=1}^{x}{1 \over 3}i>0$ Since $\sum _{k=1}^{n}Cf(i)=C\sum _{k=1}^{n}f(i)$, and $f(i)$ is arithmetic, it must be that case that $\sum _{i=1}^{x}{1 \over 3}i={1 \over 3}\cdot {1+x \over 2}x={1 \over 3}\cdot {x(x+1) \over 2}={x(x+1) \over 6}.$ From there, put the sum of the two expressions over the same denominator and simplify:
    $-500x+{x(x+1) \over 6}>0$
    $\Rightarrow {-3000x \over 6}+{x(x+1) \over 6}>0$
    $\Rightarrow {-3000x+x^{2}+x \over 6}>0$
    $\Rightarrow {-2999x+x^{2} \over 6}>0$
    $\Rightarrow -2999x+x^{2}>0$
    $\Rightarrow x(x-2999)>0$
Finally, note that the zero factor property gives $ab=0$ if and only if $a=0$ or $b=0$ or both. However, for a zero factor property in which the two multipliers give a positive product, it must be the case that either $a$ and $b$ are positive or $a$ and $b$ are negative. Through that logic, we can determine four possible solutions: $x>0$ and $x>2999$, OR $x<0$ and $x<2999$. Despite the four solutions, only one solution must be true: $x>2999$ because sigma notations only work through $x\in \mathbb {Z^{+}}$ or $x$ being the set of positive integers in the situation described in the problem. As such, all $x$ greater than $2999$ gives an acceptable solution. However, this is only the first part of the problem. Now comes the easy part.
The smallest number must be $x=3000$ in the situation described above:
    ${\begin{aligned}\log _{5}\left(\sum _{i=1}^{3,000}\left(-500+{1 \over 3}i\right)\right)&=\log _{5}\left(-500(3,000)+{3,000(3,000+1) \over 6}\right)\\&=\log _{5}\left(-1,500,000+500(3000+1)\right)\\&=\log _{5}\left(-1,500,000+1,500,000+500)\right)\\&=\log _{5}\left(500\right)\end{aligned}}$
Finally, simplify $\log _{5}\left(500\right)$ to get the final answer:
    ${\begin{aligned}\log _{5}\left(500\right)&=\log _{5}\left(100\cdot 5\right)\\&=\log _{5}\left(10^{2}\cdot 5\right)\\&=\log _{5}\left((5\cdot 2)^{2}\cdot 5\right)\\&=\log _{5}\left(5^{2}\cdot 2^{2}\cdot 5\right)\\&=\log _{5}\left(2^{2}\cdot 5^{3}\right)\\&=3\log _{5}\left(5\right)+2\log _{5}\left(2\right)+1\\&=3+2\log _{5}\left(2\right)\blacksquare \end{aligned}}$
$58\times 1$
$xy$
$3$
$1$
$1$
$3$
$58\times 1$
    $A(20)=\sum _{i=1}^{19}\left(-500+{1 \over 3}i\right)\cdot \sum _{i=1}^{19}\left(i\right)=5900$
Notice the why the above answer is wrong. Multiplying the products of two sums does not imply the product is the sum of each index. It is the product of the sum of all the index with the given functions. The student that realizes this mistake will soon realize this problem is harder than it is.
The problem can be easily fixed if one fixes the parentheses. Recall how we always state to pay attention to parentheses. This is one big reason why. The student that made the mistake above realizes a trick but did not utilize it correctly. As a result, the above answer is an over-estimate of the correctly applied solution.
Let the length of each index be $l(i)$ and the width be $w(i)$. Analyzing the pattern for this above process gives us insight into the resulting area of the cumulative area for each index. We want to find the end result, some $k$, that allows the width to be the following:
    $1\leq l(k)\leq 3$.
First, notice how function $l(i)$ is arithmetic and decreasing. We start at $l(1)=58$, and the difference is $3$, so
    $l(i)=58-3(i-1)$.
The same is true for $w(i)$. Because the width starts at $i=1$ and ends at $i=k$,
    $w(i)=i$
Given we know $1\leq l(k)\leq 3$,
    $1\leq 58-3(k-1)\leq 3$
    $-57\leq -3(k-1)\leq -55$
    $19\leq k-1\leq {\frac {55}{3}}$
    $20\leq k\leq {\frac {55}{3}}+1$
For $l(k)\in \mathbb {Z} ^{+}$, $k=20$. Therefore, $w(20)$ is also a maximum. The cumulative area function is given by:
    $A(k)=\sum _{i=1}^{k}\left[w(i)\cdot l(i)\right]$
All that is left to do is apply the formulas and use a calculator to find the answer:
    ${\begin{aligned}A(k)&=\sum _{i=1}^{k}\left[(i)\cdot \left(58-3(i-1)\right)\right]\\&=\sum _{i=1}^{k}\left[58i-3i(i-1)\right]\\&=\sum _{i=1}^{k}\left(58i-3i^{2}-3i\right)\\&=\sum _{i=1}^{k}\left(55i-3i^{2}\right)\end{aligned}}$
    $A(20)=2940{\text{ units}}^{2}\blacksquare$
    $A(20)=\sum _{i=1}^{19}\left(-500+{1 \over 3}i\right)\cdot \sum _{i=1}^{19}\left(i\right)=5900$
Notice the why the above answer is wrong. Multiplying the products of two sums does not imply the product is the sum of each index. It is the product of the sum of all the index with the given functions. The student that realizes this mistake will soon realize this problem is harder than it is.
The problem can be easily fixed if one fixes the parentheses. Recall how we always state to pay attention to parentheses. This is one big reason why. The student that made the mistake above realizes a trick but did not utilize it correctly. As a result, the above answer is an over-estimate of the correctly applied solution.
Let the length of each index be $l(i)$ and the width be $w(i)$. Analyzing the pattern for this above process gives us insight into the resulting area of the cumulative area for each index. We want to find the end result, some $k$, that allows the width to be the following:
    $1\leq l(k)\leq 3$.
First, notice how function $l(i)$ is arithmetic and decreasing. We start at $l(1)=58$, and the difference is $3$, so
    $l(i)=58-3(i-1)$.
The same is true for $w(i)$. Because the width starts at $i=1$ and ends at $i=k$,
    $w(i)=i$
Given we know $1\leq l(k)\leq 3$,
    $1\leq 58-3(k-1)\leq 3$
    $-57\leq -3(k-1)\leq -55$
    $19\leq k-1\leq {\frac {55}{3}}$
    $20\leq k\leq {\frac {55}{3}}+1$
For $l(k)\in \mathbb {Z} ^{+}$, $k=20$. Therefore, $w(20)$ is also a maximum. The cumulative area function is given by:
    $A(k)=\sum _{i=1}^{k}\left[w(i)\cdot l(i)\right]$
All that is left to do is apply the formulas and use a calculator to find the answer:
    ${\begin{aligned}A(k)&=\sum _{i=1}^{k}\left[(i)\cdot \left(58-3(i-1)\right)\right]\\&=\sum _{i=1}^{k}\left[58i-3i(i-1)\right]\\&=\sum _{i=1}^{k}\left(58i-3i^{2}-3i\right)\\&=\sum _{i=1}^{k}\left(55i-3i^{2}\right)\end{aligned}}$
    $A(20)=2940{\text{ units}}^{2}\blacksquare$

### Infinite Geometric Series

There is this old and famous paradox that perplexed many who listened to it. The problem stems from the Greek thinker Zeno of Elea. This old adage was the first example of reductio ad absurdum (disproving a statement by showing that the application of the statement will lead to a contradiction, and so the original statement cannot be true). However, today, a version is no longer a paradox, and it took the invention of calculus to prove this enigma is not an enigma$^{[seefootnote3]}$.
The version of the problem that demonstrates the enigma is proven and shown below.
Note: the paradox will be shown soon.
The age-old questions of most students in mathematics arise: "When are we going to need this?" The next exploration gives one problem in which everything you have learned will be tested. This is definitely harder than many CLEP exam questions; however, problems are what make you better at math. You learn as you do, and so you must.
(a) $10{\sqrt {10{\sqrt {10{\sqrt {10{\sqrt {10{\sqrt {10{\sqrt {\ldots }}}}}}}}}}}}$ (b) $10{\sqrt[{n}]{10{\sqrt[{n}]{10{\sqrt[{n}]{10{\sqrt[{n}]{10{\sqrt[{n}]{10{\sqrt[{n}]{\ldots }}}}}}}}}}}}$
    (i) Given general form $a{\sqrt[{n}]{a{\sqrt[{n}]{a{\sqrt[{n}]{a{\sqrt[{n}]{a{\sqrt[{n}]{a{\sqrt[{n}]{\ldots }}}}}}}}}}}}$ of expression (b), find at least two integers, $a$ and $n$, that will make the general expression equal to a perfect cube number. If it is impossible, prove that it is not possible.
(c) $3{\sqrt {5{\sqrt {3{\sqrt {5{\sqrt {3{\sqrt {5{\sqrt {3{\sqrt {\ldots }}}}}}}}}}}}}}$
    (i) Given general form $a{\sqrt {b{\sqrt {a{\sqrt {b{\sqrt {a{\sqrt {b{\sqrt {a{\sqrt {\ldots }}}}}}}}}}}}}}$ of expression (c), prove there are infinitely many distinct integers $a$ and $b$ that will make the general expression equal to a perfect square. Give at least one example.
(a)
Since each term must converge to some finite value, and since each term in the sequence is multiplied by the ${\sqrt {10}}$, it is best to think of this problem as if were one term in the sequence. Discover the pattern and you may evaluate each series. Let $x_{1}=10$. Note that
    ${\begin{aligned}x_{2}&=10{\sqrt {x_{1}}}\\&=10{\sqrt {10}}\\&=10\cdot 10^{1 \over 2}\\&=10^{1+{1 \over 2}}\end{aligned}}$
If we keep the pattern going for each term, we find that $x_{n}=10{\sqrt {x_{n-1}}}=10^{1+{1 \over 2}+{1 \over 4}+\ldots +{1 \over 2^{n}}}$. This is simply a recursive geometric sequence, so in effect, as we keep adding terms to infinity in the exponent, $10^{1+{1 \over 2}+{1 \over 4}+\ldots +{1 \over 2^{n}}+\ldots }$. Note that $1+{1 \over 2}+{1 \over 4}+\ldots$ is an infinite geometric series, so we may use the summation formula:
    $\sum _{i=1}^{\infty }\left(ar^{i-1}\right)={a \over 1-r}$. Therefore, because ${1 \over 1-{1 \over 2}}=2$, the following is also true: $10^{2}=100$. As such,
$10{\sqrt {10{\sqrt {10{\sqrt {10{\sqrt {10{\sqrt {10{\sqrt {\ldots }}}}}}}}}}}}=100$
Note: there is another way to "solve" these type of expressions; however, the method does not "prove" the general convergence towards a number and simply denotes a probable solution(s) to the problem. Using the method above will demonstrate clear convergence rather than the other method.
(b)
The same idea for (a) is used. The only difference will be that the sequence denotes a general pattern for a given term inside a square root. Note that first term $a=10$ and second term $a_{2}=10{\sqrt[{n}]{10}}$. Given ${\sqrt[{k}]{a^{m}}}=a^{m \over k}$, the following must be true: $a_{2}=10^{1+{\frac {1}{n}}}$. If we keep the pattern going, we find the following is true: $a_{L}=10^{1+{\frac {1}{n}}+{\frac {1}{n^{2}}}+\ldots +{\frac {1}{n^{L-1}}}}$. As you keep going towards infinity, the power has an infinite geometric series that has
    $\sum _{i=1}^{\infty }\left(ar^{i-1}\right)={a \over 1-r}={1 \over 1-{\frac {1}{n}}}$. Note: ${1 \over 1-{\frac {1}{n}}}={1 \over {\frac {n}{n}}-{\frac {1}{n}}}={1 \over {\frac {n-1}{n}}}={\frac {n}{n-1}}$, the following is also true: $10^{\frac {n}{n-1}}$. As such,
$10{\sqrt[{n}]{10{\sqrt[{n}]{10{\sqrt[{n}]{10{\sqrt[{n}]{10{\sqrt[{n}]{10{\sqrt[{n}]{\ldots }}}}}}}}}}}}=10^{\frac {n}{n-1}}.$
    (i)
    Let $x\in \mathbb {Z} ^{+}$ be a perfect cubed number $u^{3}=x$ for $u\in \mathbb {Z} ^{+}$. If $a{\sqrt[{n}]{a{\sqrt[{n}]{a{\sqrt[{n}]{a{\sqrt[{n}]{a{\sqrt[{n}]{a{\sqrt[{n}]{\ldots }}}}}}}}}}}}=x=u^{3}$, then for $a{\sqrt[{n}]{a{\sqrt[{n}]{a{\sqrt[{n}]{a{\sqrt[{n}]{a{\sqrt[{n}]{a{\sqrt[{n}]{\ldots }}}}}}}}}}}}=a^{\frac {n}{n-1}}$, $a=u$.
    Given this information,
       ${\frac {n}{n-1}}=3\Leftrightarrow 3(n-1)=n$
       $\Leftrightarrow 3n-3=n\Leftrightarrow n={\frac {2}{3}}$
    Because $n$ does not belong to the set of integers, there can be no possible solution for this given problem. $\blacksquare$
(c)
Again, a similar idea is used as in (a) for solving the expression, although this one may be the most difficult one. Here, the difference is you cannot use simply $3$ because $5$ is inside the square root. Instead, use $x_{1}=3{\sqrt {5}}$. To continue the second sequence, $x_{2}=3{\sqrt {5{\sqrt {x_{1}}}}}$. In fact, the recursive formula continues as such: $x_{n}=3{\sqrt {5{\sqrt {x_{n-1}}}}}$. Focus on $x_{2}$:
    $x_{2}=3{\sqrt {5{\sqrt {3{\sqrt {5}}}}}}=3\left(5\left[3\left(5\right)^{\frac {1}{2}}\right]^{\frac {1}{2}}\right)^{\frac {1}{2}}=3\cdot 5^{\frac {1}{8}}\cdot 3^{\frac {1}{4}}\cdot 5^{\frac {1}{2}}$
Because multiplying is commutative, multiply all terms involving the same base to the other and group it like so:
    $3\cdot 5^{\frac {1}{8}}\cdot 3^{\frac {1}{4}}\cdot 5^{\frac {1}{2}}=\left(3\cdot 3^{\frac {1}{4}}\right)\cdot \left(5^{\frac {1}{2}}\cdot 5^{\frac {1}{8}}\right)=\left(3^{1+{\frac {1}{4}}}\right)\left(5^{{\frac {1}{2}}+{\frac {1}{8}}}\right)$.
From here, it is very easy to see that the expression $\left(3^{1+{\frac {1}{4}}}\right)\left(5^{{\frac {1}{2}}+{\frac {1}{8}}}\right)$ will on the next pattern be $\left(3^{1+{\frac {1}{4}}+{\frac {1}{16}}}\right)\left(5^{{\frac {1}{2}}+{\frac {1}{8}}+{\frac {1}{32}}}\right)$. This is simply because the values alternate inside the given square root. As the value of one is alternated, the value of each square root increases by ${\frac {1}{2}}$ per number. Since a multiple of each number does not care about the order by which it is multiplied (i.e. multiplication is commutative), place the common base multipliers next to the other to get what we found. Therefore, a pattern emerges that can be exploited: $\left(3^{f(n)}\right)\left(5^{g(n)}\right){\text{, where }}f(n)=\sum _{n=1}^{k}\left({\frac {1}{4}}\right)^{n}{\text{ and }}g(n)=\sum _{n=1}^{k}\left({\frac {1}{2}}\left[{\frac {1}{4}}\right]^{n}\right).$
Let $k$ tend towards infinity. Evaluate each function to find the final value:
    $f(n)=\sum _{n=1}^{\infty }\left({\frac {1}{4}}\right)^{n}={\frac {1}{\frac {3}{4}}}={\frac {4}{3}}.$
    $g(n)=\sum _{n=1}^{\infty }\left({\frac {1}{2}}\left[{\frac {1}{4}}\right]^{n}\right)={\frac {\frac {1}{2}}{\frac {3}{4}}}={\frac {2}{3}}.$
Therefore, the final value for (c) must be $3{\sqrt {5{\sqrt {3{\sqrt {5{\sqrt {3{\sqrt {5{\sqrt {3{\sqrt {\ldots }}}}}}}}}}}}}}=\left(3^{\frac {4}{3}}\right)\left(5^{\frac {2}{3}}\right).$
    (i)
    As can be easily be inferred from item (c), $a{\sqrt {b{\sqrt {a{\sqrt {b{\sqrt {a{\sqrt {b{\sqrt {a{\sqrt {\ldots }}}}}}}}}}}}}}=a^{\frac {4}{3}}b^{\frac {2}{3}}$. To make the result a positive integer, where $a^{\frac {4}{3}}b^{\frac {2}{3}}\in \mathbb {Z} ^{+}$, it needs to be the case that $a=u^{3}$ and $b=v^{3}$, for $u,v\in \mathbb {Z} ^{+}$.
    Let $a^{\frac {4}{3}}b^{\frac {2}{3}}=t^{2}\in \mathbb {Z} ^{+}$. Substituting for information we know,
       $u^{4}v^{2}=t^{2}$.
    Let $u^{4}=s^{2}$.
       $s^{2}v^{2}=t^{2}\Leftrightarrow (sv)^{2}=t^{2}\Rightarrow sv=t$.
    There exists some $a,b\in \mathbb {Z} ^{+}$ that allows the expression to be square. Because a value simply needs to be cube for each, there exists infinitely many possibilities. $\blacksquare$ One non-trivial example would be $(a,b)=(29\,791,887\,503\,681)$, which will result in the perfect square $8.52891037\cdot 10^{11}$, for which the square root gives $923\,521$. Taking the square root further gives $961$; furthering such actions gives $31$.
(a)
Since each term must converge to some finite value, and since each term in the sequence is multiplied by the ${\sqrt {10}}$, it is best to think of this problem as if were one term in the sequence. Discover the pattern and you may evaluate each series. Let $x_{1}=10$. Note that
    ${\begin{aligned}x_{2}&=10{\sqrt {x_{1}}}\\&=10{\sqrt {10}}\\&=10\cdot 10^{1 \over 2}\\&=10^{1+{1 \over 2}}\end{aligned}}$
If we keep the pattern going for each term, we find that $x_{n}=10{\sqrt {x_{n-1}}}=10^{1+{1 \over 2}+{1 \over 4}+\ldots +{1 \over 2^{n}}}$. This is simply a recursive geometric sequence, so in effect, as we keep adding terms to infinity in the exponent, $10^{1+{1 \over 2}+{1 \over 4}+\ldots +{1 \over 2^{n}}+\ldots }$. Note that $1+{1 \over 2}+{1 \over 4}+\ldots$ is an infinite geometric series, so we may use the summation formula:
    $\sum _{i=1}^{\infty }\left(ar^{i-1}\right)={a \over 1-r}$. Therefore, because ${1 \over 1-{1 \over 2}}=2$, the following is also true: $10^{2}=100$. As such,
$10{\sqrt {10{\sqrt {10{\sqrt {10{\sqrt {10{\sqrt {10{\sqrt {\ldots }}}}}}}}}}}}=100$
Note: there is another way to "solve" these type of expressions; however, the method does not "prove" the general convergence towards a number and simply denotes a probable solution(s) to the problem. Using the method above will demonstrate clear convergence rather than the other method.
(b)
The same idea for (a) is used. The only difference will be that the sequence denotes a general pattern for a given term inside a square root. Note that first term $a=10$ and second term $a_{2}=10{\sqrt[{n}]{10}}$. Given ${\sqrt[{k}]{a^{m}}}=a^{m \over k}$, the following must be true: $a_{2}=10^{1+{\frac {1}{n}}}$. If we keep the pattern going, we find the following is true: $a_{L}=10^{1+{\frac {1}{n}}+{\frac {1}{n^{2}}}+\ldots +{\frac {1}{n^{L-1}}}}$. As you keep going towards infinity, the power has an infinite geometric series that has
    $\sum _{i=1}^{\infty }\left(ar^{i-1}\right)={a \over 1-r}={1 \over 1-{\frac {1}{n}}}$. Note: ${1 \over 1-{\frac {1}{n}}}={1 \over {\frac {n}{n}}-{\frac {1}{n}}}={1 \over {\frac {n-1}{n}}}={\frac {n}{n-1}}$, the following is also true: $10^{\frac {n}{n-1}}$. As such,
$10{\sqrt[{n}]{10{\sqrt[{n}]{10{\sqrt[{n}]{10{\sqrt[{n}]{10{\sqrt[{n}]{10{\sqrt[{n}]{\ldots }}}}}}}}}}}}=10^{\frac {n}{n-1}}.$
    (i)
    Let $x\in \mathbb {Z} ^{+}$ be a perfect cubed number $u^{3}=x$ for $u\in \mathbb {Z} ^{+}$. If $a{\sqrt[{n}]{a{\sqrt[{n}]{a{\sqrt[{n}]{a{\sqrt[{n}]{a{\sqrt[{n}]{a{\sqrt[{n}]{\ldots }}}}}}}}}}}}=x=u^{3}$, then for $a{\sqrt[{n}]{a{\sqrt[{n}]{a{\sqrt[{n}]{a{\sqrt[{n}]{a{\sqrt[{n}]{a{\sqrt[{n}]{\ldots }}}}}}}}}}}}=a^{\frac {n}{n-1}}$, $a=u$.
    Given this information,
       ${\frac {n}{n-1}}=3\Leftrightarrow 3(n-1)=n$
       $\Leftrightarrow 3n-3=n\Leftrightarrow n={\frac {2}{3}}$
    Because $n$ does not belong to the set of integers, there can be no possible solution for this given problem. $\blacksquare$
(c)
Again, a similar idea is used as in (a) for solving the expression, although this one may be the most difficult one. Here, the difference is you cannot use simply $3$ because $5$ is inside the square root. Instead, use $x_{1}=3{\sqrt {5}}$. To continue the second sequence, $x_{2}=3{\sqrt {5{\sqrt {x_{1}}}}}$. In fact, the recursive formula continues as such: $x_{n}=3{\sqrt {5{\sqrt {x_{n-1}}}}}$. Focus on $x_{2}$:
    $x_{2}=3{\sqrt {5{\sqrt {3{\sqrt {5}}}}}}=3\left(5\left[3\left(5\right)^{\frac {1}{2}}\right]^{\frac {1}{2}}\right)^{\frac {1}{2}}=3\cdot 5^{\frac {1}{8}}\cdot 3^{\frac {1}{4}}\cdot 5^{\frac {1}{2}}$
Because multiplying is commutative, multiply all terms involving the same base to the other and group it like so:
    $3\cdot 5^{\frac {1}{8}}\cdot 3^{\frac {1}{4}}\cdot 5^{\frac {1}{2}}=\left(3\cdot 3^{\frac {1}{4}}\right)\cdot \left(5^{\frac {1}{2}}\cdot 5^{\frac {1}{8}}\right)=\left(3^{1+{\frac {1}{4}}}\right)\left(5^{{\frac {1}{2}}+{\frac {1}{8}}}\right)$.
From here, it is very easy to see that the expression $\left(3^{1+{\frac {1}{4}}}\right)\left(5^{{\frac {1}{2}}+{\frac {1}{8}}}\right)$ will on the next pattern be $\left(3^{1+{\frac {1}{4}}+{\frac {1}{16}}}\right)\left(5^{{\frac {1}{2}}+{\frac {1}{8}}+{\frac {1}{32}}}\right)$. This is simply because the values alternate inside the given square root. As the value of one is alternated, the value of each square root increases by ${\frac {1}{2}}$ per number. Since a multiple of each number does not care about the order by which it is multiplied (i.e. multiplication is commutative), place the common base multipliers next to the other to get what we found. Therefore, a pattern emerges that can be exploited: $\left(3^{f(n)}\right)\left(5^{g(n)}\right){\text{, where }}f(n)=\sum _{n=1}^{k}\left({\frac {1}{4}}\right)^{n}{\text{ and }}g(n)=\sum _{n=1}^{k}\left({\frac {1}{2}}\left[{\frac {1}{4}}\right]^{n}\right).$
Let $k$ tend towards infinity. Evaluate each function to find the final value:
    $f(n)=\sum _{n=1}^{\infty }\left({\frac {1}{4}}\right)^{n}={\frac {1}{\frac {3}{4}}}={\frac {4}{3}}.$
    $g(n)=\sum _{n=1}^{\infty }\left({\frac {1}{2}}\left[{\frac {1}{4}}\right]^{n}\right)={\frac {\frac {1}{2}}{\frac {3}{4}}}={\frac {2}{3}}.$
Therefore, the final value for (c) must be $3{\sqrt {5{\sqrt {3{\sqrt {5{\sqrt {3{\sqrt {5{\sqrt {3{\sqrt {\ldots }}}}}}}}}}}}}}=\left(3^{\frac {4}{3}}\right)\left(5^{\frac {2}{3}}\right).$
    (i)
    As can be easily be inferred from item (c), $a{\sqrt {b{\sqrt {a{\sqrt {b{\sqrt {a{\sqrt {b{\sqrt {a{\sqrt {\ldots }}}}}}}}}}}}}}=a^{\frac {4}{3}}b^{\frac {2}{3}}$. To make the result a positive integer, where $a^{\frac {4}{3}}b^{\frac {2}{3}}\in \mathbb {Z} ^{+}$, it needs to be the case that $a=u^{3}$ and $b=v^{3}$, for $u,v\in \mathbb {Z} ^{+}$.
    Let $a^{\frac {4}{3}}b^{\frac {2}{3}}=t^{2}\in \mathbb {Z} ^{+}$. Substituting for information we know,
       $u^{4}v^{2}=t^{2}$.
    Let $u^{4}=s^{2}$.
       $s^{2}v^{2}=t^{2}\Leftrightarrow (sv)^{2}=t^{2}\Rightarrow sv=t$.
    There exists some $a,b\in \mathbb {Z} ^{+}$ that allows the expression to be square. Because a value simply needs to be cube for each, there exists infinitely many possibilities. $\blacksquare$ One non-trivial example would be $(a,b)=(29\,791,887\,503\,681)$, which will result in the perfect square $8.52891037\cdot 10^{11}$, for which the square root gives $923\,521$. Taking the square root further gives $961$; furthering such actions gives $31$.

#### Proof of Infinite Geometric Series

$\sum _{i=1}^{\infty }\left(ar^{i-1}\right)={a \over 1-r}{\text{ if }}0<\left|r\right\vert <1$. How do we prove this is true? Again, this is way above most curriculum required textbooks. However, the goal of many mathematics textbooks is to get the user inspired in math and to use those concepts into other fields of interest. No matter what given field a person is going into, they are going to need to solve problems: a history major needs to understand what a person meant; an English major needs to find the right words that help them convey their ideas; a Science major will have to apply mathematical concepts to communicate ideas about science. As such, learn to love problems, for conquering a problem will make you stronger.
Note: what you are about to learn is NOT required for the curriculum. If you do not understand these concepts, do not worry, for it does not matter for the CLEP exam. These proofs are only to build a mathematical understanding of these concepts. As such, you may skip these if you want.
Prerequisite for the proof: Imagine we want to find values of some function $f(x)$. Define this function as $f(x)={x^{2}-4 \over x-2}$. If $x=2$, then $f(2)={(2)^{2}-4 \over (2)-2}={0 \over 0}$. Well, it seems this value is undefined when $x=2$, so it seems it does not equal anything. However, let's try graphing this rational function.
First, let us divide: ${x^{2}-4 \over x-2}={(x-2)(x+2) \over x-2}=x+2$. It seems once we simplify the rational expression, we are left with $x+2$. However, This is not exactly true because our function is still undefined at $x=2$, so we must add a constraint: $f(x)=x+2{\text{ if }}x\neq 2$. To put it more formally:
$f(x)={\begin{cases}x+2,&{\text{if}}&-\infty <x<2\\{x^{2}-4 \over x-2},&{\text{if}}&x=2\\x+2,&{\text{if}}&2<x<\infty \end{cases}}$
If we graph the relation, we find that there is a "hole" at that point in the function. Let's use values close to 2 so that we abide by those rules.
${\begin{array}{|c|c|}x&f(x)\\\hline 1.9&(1.9)+2=3.9\\\hline 1.99&(1.99)+2=3.99\\\hline 1.999&(1.999)+2=3.999\\\hline 2&{\text{undefined}}\\\hline 2.001&(2.001)+2=4.001\\\hline 2.01&(2.01)+2=4.01\\\hline 2.1&(2.1)+2=4.1\end{array}}\!$
It becomes clear that as values get "closer and closer" to $2$, $f(2)$ "approaches" $4$. What is "close"? How close are your eyes to the screen? We would most likely say, "it is 45 cm. away from the screen." In which case, let's use the distance (i.e. absolute value) to define the word "close": $|2.001-2|=0.001$ and $|1.999-2|=0.001$. (Answer this question: why did we use absolute values?)
Let's define this process using two variables $x_{1}$ and $x$: If $0<\left|x_{1}-x\right\vert <n$, then those two values are "close" or "approaches $L$ of function $f(x)$," where $n$ can be any value you want it to be, as long as it tolerates our definition. We can conclude the following: $\left|f(x)-L\right\vert$ is "close" when $\left|x_{1}-x\right\vert$ is close, given that $0<\left|f(x)-L\right\vert <n_{f}$. If we think about it, we have sufficiently proven what "close" means. Let us put this definition$^{[seefootnote4]}$ up front:
For any $0<\left|f(x)-L\right\vert <n_{f}$, there exists a $0<\left|x_{1}-x\right\vert <n$.
Now, as mentioned, mathematicians like to "work smarter, not harder," so a mathematician does not want to write out "as $x$ approaches $2$, $f(x)$ approaches $4$." Instead, we write out using our fancy notation: $\lim _{x\to 2}f(x)=4$. This $\lim$ is called "limit." All this notation says is that "as we limit our values of $x$ approaching ($\to$) $2$, we find that $f(x)$ approaches $4$." Now, you have a sufficient foundation for limits and are ready to begin the proof.
Read the prerequisite to understand this proof.
If we want to know what happens as $k$ approaches infinity ($\infty$), we can write it like this: $\lim _{k\to \infty }\sum _{i=1}^{k}\left(ar^{i-1}\right)={a\left(r^{k}-1\right) \over r-1}$
Note: ${a\left(r^{k}-1\right) \over r-1}={ar^{k}-a \over r-1}$.
What happens when $|r|>1$? For $\lim _{k\to \infty }\sum _{i=1}^{k}\left(ar^{i-1}\right)={ar^{k}-a \over r-1}$, the numerator will have values $ar^{k}$ get closer and closer to infinity as $r$ and $k$ get larger, and therefore, do not converge to a finite value.
What happens when $|r|=1$? For $\lim _{k\to \infty }\sum _{i=1}^{k}\left(ar^{i-1}\right)={ar^{k}-a \over r-1}$, the denominator will have values $(1)-1=0$, which will ${a-a \over 0}={0 \over 0}$. It is indeterminate, and there are no tricks to find a value that it will converge to. As such, when $|r|=1$, the limit is indeterminate.
What happens when $0\leq |r|<1$? For $\lim _{k\to \infty }\sum _{i=1}^{k}\left(ar^{i-1}\right)={ar^{k}-a \over r-1}$, the numerator will have values $ar^{k}$ approach $0$ as $r$ and $k$ get larger. Plus, the denominator will never be $0$, so everything works out. If $\lim _{k\to \infty }ar^{k}=0$, given that $0\leq |r|<1$, then $\lim _{k\to \infty }\sum _{i=1}^{k}\left(ar^{i-1}\right)={-a \over r-1}$.
Note: ${-a \over r-1}={-a \over -(-r+1)}={a \over -r+1}={a \over 1-r}$. Ergo: $\lim _{k\to \infty }\sum _{i=1}^{k}\left(ar^{i-1}\right)={a \over 1-r}{\text{ if }}0\leq \left|r\right\vert <1.$
Note that by proving the following fact for the limit, we also "prove" the summation representation$^{[seefootnote5]}$.

## Footnotes

* The WikiBooks simply refers the formula used as the "Gaussian method" because of a famous mathematical tale. There was this intelligent mathematician named Carl Friedrich Gauss who, in second grade, was asked to find the sum of $1+2+3+4+\ldots +99+100$. The teacher that assigned this problem simply wanted some peace and quiet. However, Gauss was able to come up the answer using the same method in Example 1.3.1.a. Legend says that he did not write the method and did all the calculations in his head; he handed the piece of paper with just the answer. Nevertheless, the teacher had to check to see if he was right. It turns out, he was indeed right. If the reader needs to refer to this formula to other mathematicians, simply call it the "sum of an arithmetic series formula," although it does not have the same ring to it as the previous name.
* It is very common for many textbooks to use the following formula for a geometric series: $S_{g}={a\left(1-r^{k}\right) \over 1-r}$. Both are technically correct; however, this Wikibooks has decided to use $S_{g}={a\left(r^{k}-1\right) \over r-1}$ because it is easier to remember that you subtract the first term from the final term at index $k$.
* Although described technically correct, calculus was merely one piece of the puzzle that helped in solving the issue. The version that is described is shown in the example, but the original problem did involve calculus concepts that will not be talked about in this Wikibooks since it does not help us understand infinite geometric series.
* Note the formal definition is technically complete; however, the variables refered to the distances are a little different. Instead of $0<\left|x_{1}-x\right\vert <n$, write it as $0<\left|x_{1}-x\right\vert <\delta$ (read $\delta$ aloud as "delta"). Instead of $0<\left|f(x)-L\right\vert <n_{f}$, write it as $0<\left|f(x)-L\right\vert <\epsilon$ (read $\epsilon$ aloud as "epsilon"). Consequently, the definition given for "close" changes: "For any $0<\left|f(x)-L\right\vert <\epsilon$, there exists a $0<\left|x_{1}-x\right\vert <\delta$." Also note that the definition may be written a little different as "for any $\epsilon >0$, there is a $\delta >0$ such that $\left|f(x)-L\right\vert <\epsilon$ and $\left|x_{1}-x\right\vert <\delta$." For the interest of not confusing the reader, we decided to make the definition more simple to understand.
* Note that the limit can be used for any function. Let $L(x)=2x$. Say we want to find $\lim _{x\to 10}L(x)$. We can do that because as we get closer and closer to $10$, we find that $L(x)$ gets closer and closer to $20$. However, we can simplify the process by saying that $\lim _{x\to L}f(x)=f(L)$, given that a function is continuous. Because there are no jumps or asymptotes in $f(i)=\left(ar^{i-1}\right)$ as long as $0\leq |r|<1$, we can state $\sum _{i=1}^{\infty }f(i)={a \over 1-r}{\text{ if }}0\leq \left|r\right\vert <1$. This is why we can use the limit and say that $\lim _{k\to \infty }\sum _{i=1}^{k}\left(ar^{i-1}\right)=\sum _{i=1}^{\infty }\left(ar^{i-1}\right)={a \over 1-r}{\text{ if }}0\leq \left|r\right\vert <1$.