# Calculus/Direct Comparison Test
## Direct Comparison Test

The Direct Comparison Test (DCT) is sometimes simply called The Comparison Test. However, like we do here, many books include the word 'Direct' in the name to clearly separate this test from the Limit Comparison Test.
Some instructors will tell you that this test is very basic and intuitive, but at first it can be difficult to understand and use for many students. However, once it is mastered, it is a quite powerful test and it will work more easily than the Limit Comparison Test since we don't need to evaluate a limit.
Key - - The key is to set up the inequality correctly. This can only be done once a test series is chosen and you know the convergence/divergence of that test series.
First we present the definition of the Direct Comparison Test and then we explain each part.

### Direct Comparison Test Definition

For the series $\textstyle {\sum {a_{n}}}$ and the test series $\textstyle {\sum {t_{n}}}$ where $t_{n}>0$
Convergence To prove convergence of $\textstyle {\sum {a_{n}}}$, $\textstyle {\sum {t_{n}}}$ must converge and $0<a_{n}\leq t_{n}$
Divergence To prove divergence of $\textstyle {\sum {a_{n}}}$, $\textstyle {\sum {t_{n}}}$ must diverge and $0<t_{n}\leq a_{n}$

### Direct Comparison Test Quick Notes

| used to prove convergence | yes |
| used to prove divergence | yes |
| can be inconclusive | no |
* Notice that we do not specify the $n$-values on the sum. This is common in calculus and it just means that, for this test, it doesn't matter where the series starts (but it always 'ends' at infinity, since this is an infinite series).
* $\textstyle {\sum {a_{n}}}$ is the series of which we are trying to determine convergence or divergence and it is given in the problem statement.
* $\textstyle {\sum {t_{n}}}$ is the test series that you choose for comparison.
* In the course of using this test, you may need to find some real, finite value $N>0$ where the inequalities hold for all $n\geq N$.
* Be careful when setting up the inequality. It is set up differently depending on whether you are assuming convergence or divergence.
* A subtle distinction for using this test that is implied in the inequalities in the above definition is that $a_{n}>0$. So be careful not to use this on a series that is alternating or that contains any negative terms.
* Some instructors may say that this test can be inconclusive. This is not correct. This test will always tell you if the series $\textstyle {\sum {a_{n}}}$ either converges or diverges IF a valid test series can be found AND the inequality can be proven. There will be times when a valid test series cannot be found or when the inequality cannot be proven but this does not mean that the DCT is inconclusive.

## How To Use the Direct Comparison Test

There are three main steps to using this test.
| Step 1 | choose a test series |
| Step 2 | set up the inequality |
| Step 3 | prove the inequality holds |
The difficult thing about this test is that it seems like you are expected to already know whether the series converges or diverges before you even use the test. It helps to have a feel for it (and, with enough practice, you will develop this over time) but if you don't, you can guess. If you reach a dead-end when trying to prove the inequality holds, try to prove the other direction. You will find more suggestions on this in step 1, below.
Let's look at the details of each step.

### Step 1 - Choose A Test Series

When you are first learning this technique, it may look like the test series comes out of thin air and you just randomly choose one and see if it works. If it doesn't, you try another one. This is not the best way to choose a test series. The best way I've found is to use the series you are asked to work with and come up with the test series. There are several things to consider.
The first key is to choose a test series that you know converges or diverges AND that will help you get a finite, positive limit.
Idea 1: If you have polynomials in both the numerator and denominator of a fraction, drop all terms except for the highest power terms (in both parts) and simplify. Drop any constants. What you end up with may be a good comparison series. The reason this works is that, as $n$ gets larger and larger, the highest powers dominate. You will often end up with a p-series that you know either converges or diverges.
Idea 2: Choose a p-series or geometric series since you can tell right away whether it converges or diverges.
Idea 3: If you have a sine or cosine term, you are always guaranteed that the result is less than or equal to one and greater than or equal to negative one. If you don't have any bounds on the angle, these are the best you can do. So replace the sine or cosine term with one.
Idea 4: If you have a natural log, use the fact that $\ln(n)\leq n$ for $n\geq 1$ to replace $\ln(n)$ with $n$ or use $\ln(n)\geq 1$ for $n\geq 3$.
As you get experience with this test, it will become easier to determine a good test series. So work plenty of practice problems.
The additional thing you need to think about for the Direct Comparison Test that doesn't matter for the Limit Comparison Test is that you need to have a feel for whether the series $\textstyle {\sum {a_{n}}}$ converges or diverges. This requires a certain amount of experience, since it determines how you set up the inequality. But you can guess by looking at the test series you ended up with. If it diverges, then your series may too. Similarly, if the test series converges, then you want to test for convergence of your original series.
There is no general rule for choosing a test series but with some experience you will begin to see patterns and we will show you some examples and explain how to chose a test series (something that many/most books leave out).

### Step 2 - Set Up The Inequality

If you have a series $\textstyle {\sum {a_{n}}}$ and you choose a test series $\textstyle {\sum {t_{n}}}$ then you can set up the inequality in one of two ways:
| Convergence | Divergence |
| If you are assuming convergence, the test series must also converge and the inequality you need to prove is $0<a_{n}\leq t_{n}$. | If you are assuming divergence, the test series must diverge and you need to prove is $0<t_{n}\leq a_{n}$. |
Here is an idea on how to think about the direction of the inequality. If you think the series you are working with diverges, you want to pick a divergent test series that is SMALLER than the series you are working with. You can think about this smaller test series as 'pushing up' your series as $n$ increases and since the small series diverges, there is no way your series can converge since it is always being pushed up to infinity.
However, if you think your series converges, then you need to choose a convergent test series that is LARGER than your series. Then, you can think about the test series as always 'holding down' your series as $n$ increases and not allowing it to go off to infinity.

### Step 3 - Prove The Inequality Holds

Once you get the inequality set up, you need to prove that the inequality holds for all $n$ greater than some $N$. There are several techniques to do this depending on the inequality, one of which should work.

#### Technique 1 - Directly

In this case, set up the inequality and perform algebraic operations until you get an inequality that always holds. For example, we can show that $n\leq n+2$ by subtracting $n$ from both sides to get $0\leq 2$. This last inequality is always true.
NOTE: Be careful when squaring and taking square (even) roots. The resulting inequality may not be equal.

#### Technique 2 - Prove an inequality is always positive

If the direct way is not possible, try moving all the terms to one side leaving zero on the other side and then explain how the expression is always positive. For example, if we can get something like $0\leq \displaystyle {\frac {n+5}{n^{3}}}$ we can argue that, since $n$ is always positive, both the numerator and denominator are positive, resulting in the right side always being greater than zero. The key to remember here is that $n$ starts at zero or one and is always positive after that.
This technique also works if you can find a value $N$ such that the expression holds for all $n\geq N$. Similar to the last example, you can use this argument for the inequality $0\leq \displaystyle {\frac {n-5}{n^{3}}}$ by saying that for $n\geq 6$, the inequality holds.

#### Technique 3 - Using Slope

The third technique is to use the concept of slope and is best demonstrated with an example. Let's show that $n\geq \ln(n)$ using slope.
The first thing to remember about slope is, to find the slope, you take the derivative and the derivative is defined only on continuous functions. In our case $n$ is discrete ($n$ takes on the discrete values 1, 2, 3, 4, ... but no value in between these numbers), so we need to find continuous functions that have the same values at the discrete values. We don't care what is going on between the discrete values as long as the function is continuous. So for $n$, we can use $f(x)=x$ and for $\ln(n)$ we can use $g(x)=\ln(x)$ where $x$ is a continuous variable in both functions. Now we have continuous functions so that we can take derivatives. I know this is a fine point, but we need to get it right.
Okay, we need to show that $x\geq \ln(x)$ for all $x$ greater than some value. Let's call $f(x)=x$ and $g(x)=\ln(x)$. If we can show that $f(x)\geq g(x)$ for some specific value of $x$ and the slope of $f(x)$ is greater than the slope of $g(x)$, then $f(x)$ will always be greater than $g(x)$. The graphs will never cross and the inequality $x\geq \ln(x)$ will hold. You can see this intuitively in this graph (but you cannot use the graph to prove this).
Let's see if we can show this. First, we know that when $x=1$, $f(1)=1$ and $g(1)=0$. Since $1\geq 0$, we have established a point ($x=1$) where $f(x)\geq g(x)$. Now we will use slope to establish that the functions stay that way for all $x>1$.
Taking the derivatives, we have $f'(x)=1$ and $g'(x)=1/x$. We need to show $f'(x)\geq g'(x)$
Now, since, $x$ is always greater than or equal to $1$, then the slope of $f(x)$ is always larger than the slope of $g(x)$. This says that $f(x)$ is increasing at a faster rate than $g(x)$ and therefore will always be larger.
This shows that $x\geq \ln(x)$.
To recap, what we have done here is that we have found a point, $x=1$ where the inequality holds. Then we have used slope to say that the inequality holds for all values greater than that value.

#### What If The Inequality Doesn't Hold?

If you are unable to prove the inequality, then you either need to choose a different test series or try another test. Using the Direct Comparison Test takes practice and time to sink in before you can understand it and use it.
If the inequality doesn't hold, it doesn't mean that the Direct Comparison Test can't be used. We may just need to choose a different test series. One suggestion is to get a feel for what the series looks like is to plot the series on your calculator or use another test. Then see if you can extrapolate a smaller or larger series in order to come up with another test series.

## How NOT To Use The Direct Comparison Test

There are two main ways students might try to use the Direct Comparison Test that do not work.
1. Build a table using the first few values of $n$ to show that the inequality holds for all $n$. Be very, very careful to not use this technique. It is a pitfall that instructors watch for.
2. Set up the inequality incorrectly. The rest of this section shows this and explains why it doesn't work.
Here is how NOT to use the direct comparison test, i.e. when this test does not work when the inequality is set up incorrectly.
In one video I saw, the presentor looked at the example $\displaystyle {\sum _{n=2}^{\infty }{\frac {1}{n(\ln(n))^{2}}}}$ to show that you cannot use the direct comparison test by comparing this to $\displaystyle {\sum {\frac {1}{n^{3}}}}$ to prove convergence. When this happens, you have two choices.
1. You can choose a different test series.
2. You can try another test.
In this example, either of these choices will work.
1. Compare this series to $\displaystyle {\sum {\frac {1}{n^{2}}}}$.
2. Use the integral test or, perhaps, the limit comparison test.
No matter what test you use, this series converges.

## Study Tip

When learning this test, it may help you to draw graphs of what is going on. Although this is a good technique to use in general, it will especially help you with this test, since the inequalities can best be shown in a graph. You do not even need specific functions. You can just draw generic functions that are above and below a test function.
Be careful! Plotting and plugging in values is not a valid way to prove convergence or divergence of ANY series. Instructors anticipate this and will often put questions on exams that give incorrect answers this way.

## Video Recommendations

If you want a complete lecture on the Direct Comparison Test, this is a good video clip. Notice he calls this The Comparison Test, leaving out the word Direct. You need to watch only the first 40 minutes and 12 seconds. After that he discusses the Limit Comparison Test.
Here is a quick video explaining this test.
The first five and a half minutes of this video explains the direct comparison test very well.
The first two minutes of this video also contains a good explanation of the direct comparison test.
This video clip is important to watch since it shows a pitfall that almost every student falls into when using this test and most teachers will watch for.

## Application To Improper Integrals

Even if you have not had improper integrals yet, this video is excellent to watch anyway to help you visualize the direct comparison test. You don't need to understand improper integrals to get a lot out of this video.

## Example Set 1 - Test For Divergence

Use the divergent and monotonic harmonic series $\sum _{n=1}^{\infty }{\frac {1}{n}}$ to determine if $\textstyle {\sum {a_{n}}}$ is divergent, if possible, for each of following series.

### Example 1.1

$a_{n}={\frac {1}{n-1}}$
Since we are assuming divergence, the inequality we need to set up is $0<t_{n}\leq a_{n}$ where $a_{n}={\frac {1}{n-1}}$ and $t_{n}={\frac {1}{n}}$. Since $n>0$, $t_{n}=1/n$ is also greater than zero and therefore the left half of the inequality holds. so we just need to show the right half of the inequality.
This last inequality holds for all $n$. Therefore the series diverges by the DCT.
This is a great example of how to choose a test series. Constants become negligible as $n$ becomes very large. So sometimes choosing a test series by dropping constants work. Of course, this will not work all the time, which the next example shows.

### Example 1.2

$a_{n}={\frac {1}{n+1}}$
As in the previous example, we only need to show that the right half of the inequality holds.
However, this is never true so this test series cannot be used. (Some instructors might say this means that the DCT is inconclusive but actually, this is not a valid test series to prove divergence.) The test series with $t_{n}={\frac {1}{2n}}$ would be a better test for comparison.

### Example 1.3

$a_{n}={\frac {3}{n}}$
As in the previous two examples, we only need to show that the right half of the inequality holds.
The last inequality holds for all $n$. So the series diverges by the DCT.

### Example 1.4

$a_{n}={\frac {1}{\sqrt {n}}}$
As in the previous examples, we only need to show that the right half of the inequality holds.
The last inequality holds for all $n\geq 2$. So the series diverges by the DCT.
Two comments are in order here.
* Notice that we needed to find a value $N=2$ where the inequality held for all values $n\geq N$.
* Although in general it is not a good idea to square both sides of an inequality and assume it is a valid operation, it works here since $n$ is always greater than one.

### Example 1.5

$a_{n}={\frac {1}{n^{2}}}$
As in the previous examples, we only need to show that the right half of the inequality holds.
The last inequality is not true for all $n$ greater than some $N$, i.e. once $n$ gets larger than 1, the last inequality no longer holds. Therefore, this test series cannot be used to prove divergence using the DCT.
Note that the DCT can still be used on this series but since the series converges, another test series is required if we want to use the DCT.

## Example Set 2 - Test for Convergence

Use the convergent and monotonic series $\sum _{n=1}^{\infty }{\frac {1}{2^{n}}}$ to determine if $\textstyle {\sum {a_{n}}}$ is convergent, if possible, for each of following series.

### Example 2.1

$a_{n}={\frac {1}{n^{2}}}$
$n^{-2}$ decreases at a faster rate than $2^{-n}$. However, these series do not satisfy the $0\leq z_{n}\leq s_{n}$ requirement, because $\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}$ is larger than $\sum _{n=1}^{\infty }{\frac {1}{2^{n}}}$ when $n<2$. We can solve this issue by taking removing the first term from the both series to obtain $1+\sum _{n=2}^{\infty }{\frac {1}{n^{2}}}$ and $2+\sum _{n=2}^{\infty }{\frac {1}{2^{n}}}$. Now, comparing $\sum _{n=2}^{\infty }{\frac {1}{n^{2}}}$ with $\sum _{n=2}^{\infty }{\frac {1}{2^{n}}}$ shows that $\sum _{n=2}^{\infty }{\frac {1}{n^{2}}}$ is indeed convergent. Because this is convergent, adding the original $1$ will not change whether it is convergent or not, it will add $1$ to the value of convergence.

### Example 2.2

$a_{n}=e^{-x}$
$\sum _{n=1}^{\infty }{e^{-x}}$ is smaller $\sum _{n=1}^{\infty }{\frac {1}{2^{n}}}$ for every $n$, so this series is convergent.

### Example 2.3

$a_{n}={\frac {1}{2^{n}}}\sin ^{2}(x)$
$\sum _{n=1}^{\infty }{{\frac {1}{2^{n}}}\sin ^{2}(x)}$ is less than or equal to $\sum _{n=1}^{\infty }{\frac {1}{2^{n}}}$ and is greater than $0$ for every $n$ in the domain; this is because $\sin ^{2}(x)$ conforms to </math>\frac{1}{2^n}</math>, and the fact that $\sin(x)$ is squared implies that it will never be less than zero.

### Example 2.4

$a_{n}={\frac {2}{2^{n}}}$
$\sum _{n=1}^{\infty }{\frac {2}{2^{n}}}$ is convergent. Notice that this is just $2\times \sum _{n=1}^{\infty }{\frac {1}{2^{n}}}$, which is just $2$ multiplied by some finite number.

### Example 2.5

$a_{n}={\frac {1}{1.5^{n}}}$
$\sum _{n=1}^{\infty }{\frac {1}{1.5^{n}}}$ is greater than $\sum _{n=1}^{\infty }{\frac {1}{2^{n}}}$ for an infinite amount of $n$. So this test series cannot be used to test for convergence.

### Example 2.6

$a_{n}={\frac {(-1)^{n}}{2^{n}}}$
This series has negative terms, so the DCT cannot be used here.

## Direct Comparison Test Practice Problems
### Practice Problems with Written Solutions

Determine the convergence or divergence of these series using the Direct Comparison Test, if possible. If the DCT is inconclusive, use another test to determine convergence or divergence. Make sure to specify what test you used in your final answer.
1. $\sum _{n=2}^{\infty }{\frac {1}{n\ln(n)-1}}$
$\displaystyle {t_{n}={\frac {1}{n\ln(n)}}}$
$\displaystyle {t_{n}={\frac {1}{n\ln(n)}}}$
$\displaystyle {\sum _{n=2}^{\infty }{\frac {1}{n\ln(n)}}}$
$t_{n}<a_{n}$
$n>2$
${\begin{array}{rcl}\displaystyle {\frac {1}{n\ln(n)}}&<&\displaystyle {\frac {1}{n\ln(n)-1}}\\n\ln(n)-1&<&n\ln(n)\\-1&<&0\end{array}}$
$n$
$t_{n}<a_{n}$
$n>2$
$\displaystyle {\sum _{n=2}^{\infty }{\frac {1}{n\ln(n)}}}$
$t_{n}<a_{n}$
$n>2$
${\begin{array}{rcl}\displaystyle {\frac {1}{n\ln(n)}}&<&\displaystyle {\frac {1}{n\ln(n)-1}}\\n\ln(n)-1&<&n\ln(n)\\-1&<&0\end{array}}$
$n$
$t_{n}<a_{n}$
$n>2$
2. $\sum _{n=1}^{\infty }{\frac {(\cos(n))^{3}}{n^{2}+n+1}}$
$t_{n}=1/n^{2}$
$\cos(n)$
$|\cos(n)|$
$t_{n}=1/n^{2}$
$\cos(n)$
$|\cos(n)|$
First, let's look at the denominator polynomial $n^{2}+n+1$. As $n$ gets very large, the $n^{2}$ term will dominate the other two terms. So we will drop $n+1$ leaving only $n^{2}$ in the denominator. Now, let's look at the numerator. There is no easy way to determine what $(\cos(n))^{3}$ does as $n$ goes to infinity. Essentially, it oscillates between $-1$ and $+1$.
However, we know that $\cos(n)\leq 1$ and also $(\cos(n))^{3}\leq 1$. Now, we need one more piece to this that may not be obvious. Looking at $\cos(n)$, we need to make sure this will never be zero. (You will see why in a minute.) Can we say that? When IS $\cos(x)=0$? This is zero when $x$ is a multiple of $\pi /2$. In other words, does there exist a positive integer $k$ such that $n=k\pi /2$? No, $n=1,2,3,...$ and $\pi$ is irrational. This means that $n$ will never be a multiple of $\pi$ and, therefore, $\cos(n)$ will never be zero.
However, there is one additional detail that we need to deal with before we can use this test. Notice that both inequalities require that the $a_{n}$ terms be positive. This is an important detail. If we don't, we can't use the direct comparison test.
Fortunately, we have a theorem to help us. The absolute convergence theorem tells us that if $\textstyle {\sum {\left|a_{n}\right|}}$ converges, then $\textstyle {\sum {a_{n}}}$ also converges. So we will replace $\cos(n)$ with $\left|\cos(n)\right|$. So we will replace $(\cos(n))^{3}$ with $1$ in our test series and see what happens.
So, this gives us the p-series $\textstyle {\sum {1/n^{2}}}$ as a test series which converges. Call this series $\textstyle {\sum {t_{n}}}$ where $t_{n}=1/n^{2}$
Step 2 - Set up the inequality
Since the test series we ended up with converges, we set up the inequality as $0<a_{n}\leq t_{n}$. Now I hope you see the importance of making sure that $\cos(n)$ never evaluated to zero. If it did, the first part of the inequality $0<a_{n}$ would not hold and so we would not be able to use this test.
So the inequality that we need to prove is $\displaystyle {{\frac {(\left|\cos(n)\right|)^{3}}{n^{2}+n+1}}\leq {\frac {1}{n^{2}}}}$
Step 3 - Prove the inequality holds
${\begin{array}{rcl}\displaystyle {\frac {(\left|\cos(n)\right|)^{3}}{n^{2}+n+1}}&\leq &\displaystyle {\frac {1}{n^{2}}}\\\displaystyle {(\left|\cos(n)\right|)^{3}}&\leq &\displaystyle {\frac {n^{2}+n+1}{n^{2}}}\\\displaystyle {(\left|\cos(n)\right|)^{3}}&\leq &\displaystyle {1+1/n+1/n^{2}}\end{array}}$
$(\left|\cos(n)\right|)^{3}$
$n$
$\textstyle {\sum {a_{n}}}$
First, let's look at the denominator polynomial $n^{2}+n+1$. As $n$ gets very large, the $n^{2}$ term will dominate the other two terms. So we will drop $n+1$ leaving only $n^{2}$ in the denominator. Now, let's look at the numerator. There is no easy way to determine what $(\cos(n))^{3}$ does as $n$ goes to infinity. Essentially, it oscillates between $-1$ and $+1$.
However, we know that $\cos(n)\leq 1$ and also $(\cos(n))^{3}\leq 1$. Now, we need one more piece to this that may not be obvious. Looking at $\cos(n)$, we need to make sure this will never be zero. (You will see why in a minute.) Can we say that? When IS $\cos(x)=0$? This is zero when $x$ is a multiple of $\pi /2$. In other words, does there exist a positive integer $k$ such that $n=k\pi /2$? No, $n=1,2,3,...$ and $\pi$ is irrational. This means that $n$ will never be a multiple of $\pi$ and, therefore, $\cos(n)$ will never be zero.
However, there is one additional detail that we need to deal with before we can use this test. Notice that both inequalities require that the $a_{n}$ terms be positive. This is an important detail. If we don't, we can't use the direct comparison test.
Fortunately, we have a theorem to help us. The absolute convergence theorem tells us that if $\textstyle {\sum {\left|a_{n}\right|}}$ converges, then $\textstyle {\sum {a_{n}}}$ also converges. So we will replace $\cos(n)$ with $\left|\cos(n)\right|$. So we will replace $(\cos(n))^{3}$ with $1$ in our test series and see what happens.
So, this gives us the p-series $\textstyle {\sum {1/n^{2}}}$ as a test series which converges. Call this series $\textstyle {\sum {t_{n}}}$ where $t_{n}=1/n^{2}$
Step 2 - Set up the inequality
Since the test series we ended up with converges, we set up the inequality as $0<a_{n}\leq t_{n}$. Now I hope you see the importance of making sure that $\cos(n)$ never evaluated to zero. If it did, the first part of the inequality $0<a_{n}$ would not hold and so we would not be able to use this test.
So the inequality that we need to prove is $\displaystyle {{\frac {(\left|\cos(n)\right|)^{3}}{n^{2}+n+1}}\leq {\frac {1}{n^{2}}}}$
Step 3 - Prove the inequality holds
${\begin{array}{rcl}\displaystyle {\frac {(\left|\cos(n)\right|)^{3}}{n^{2}+n+1}}&\leq &\displaystyle {\frac {1}{n^{2}}}\\\displaystyle {(\left|\cos(n)\right|)^{3}}&\leq &\displaystyle {\frac {n^{2}+n+1}{n^{2}}}\\\displaystyle {(\left|\cos(n)\right|)^{3}}&\leq &\displaystyle {1+1/n+1/n^{2}}\end{array}}$
$(\left|\cos(n)\right|)^{3}$
$n$
$\textstyle {\sum {a_{n}}}$
3. $\sum _{n=1}^{\infty }{\frac {\sin(n)}{n^{3}+n+1}}$
$t_{n}=1/n^{3}$
$\sin(n)$
$|\sin(n)|$
$t_{n}=1/n^{3}$
$\sin(n)$
$|\sin(n)|$
$\displaystyle {a_{n}={\frac {\sin(n)}{n^{3}+n+1}}}$
Before we start with the direct comparison test, we have to look carefully at the requirements. Notice that in both inequalities, we require that $a_{n}>0$. However, for this series, the sine term introduces some negative terms. So we can't just directly use the direct comparison test. This is an important detail. If we don't handle this detail, we can't use the direct comparison test.
Fortunately, we have a theorem to help us. The absolute convergence theorem tells us that if $\sum {|a_{n}|}$ converges, then $\sum {a_{n}}$ also converges. In this case, $\displaystyle {|a_{n}|=\left|{\frac {\sin(n)}{n^{3}+n+1}}\right|={\frac {|\sin(n)|}{|n^{3}+n+1|}}}$
Since $n>0$ the denominator is always positive so we just need to replace $\sin(n)$ with $|\sin(n)|$. So, in this problem, we will work toward convergence. If we get the result that the series diverges, then we can't use the absolute convergence theorem and the result is inconclusive.
For large $n$ the $n^{3}$ term dominates. In the numerator, $|\sin(n)|$ is always less than or equal to one, let's compare this series with the test series $\sum {t_{n}}$ where $\displaystyle {t_{n}={\frac {1}{n^{3}}}}$.
Since $t_{n}$ is a p-series with $p=3>1$, the test series converges. So the direct comparison test requires us to set up the inequality as $a_{n}\leq t_{n}$.
${\begin{array}{rcl}a_{n}&\leq &t_{n}\\\displaystyle {\frac {|\sin(n)|}{n^{3}+n+1}}&\leq &\displaystyle {\frac {1}{n^{3}}}\\|\sin(n)|&\leq &\displaystyle {\frac {n^{3}+n+1}{n^{3}}}\\|\sin(n)|&\leq &\displaystyle {1+{\frac {1}{n^{2}}}+{\frac {1}{n^{3}}}}\end{array}}$
The last inequality is always true since $|\sin(n)|\leq 1$ and the right side is always greater than one. So the series converges.
$\displaystyle {a_{n}={\frac {\sin(n)}{n^{3}+n+1}}}$
Before we start with the direct comparison test, we have to look carefully at the requirements. Notice that in both inequalities, we require that $a_{n}>0$. However, for this series, the sine term introduces some negative terms. So we can't just directly use the direct comparison test. This is an important detail. If we don't handle this detail, we can't use the direct comparison test.
Fortunately, we have a theorem to help us. The absolute convergence theorem tells us that if $\sum {|a_{n}|}$ converges, then $\sum {a_{n}}$ also converges. In this case, $\displaystyle {|a_{n}|=\left|{\frac {\sin(n)}{n^{3}+n+1}}\right|={\frac {|\sin(n)|}{|n^{3}+n+1|}}}$
Since $n>0$ the denominator is always positive so we just need to replace $\sin(n)$ with $|\sin(n)|$. So, in this problem, we will work toward convergence. If we get the result that the series diverges, then we can't use the absolute convergence theorem and the result is inconclusive.
For large $n$ the $n^{3}$ term dominates. In the numerator, $|\sin(n)|$ is always less than or equal to one, let's compare this series with the test series $\sum {t_{n}}$ where $\displaystyle {t_{n}={\frac {1}{n^{3}}}}$.
Since $t_{n}$ is a p-series with $p=3>1$, the test series converges. So the direct comparison test requires us to set up the inequality as $a_{n}\leq t_{n}$.
${\begin{array}{rcl}a_{n}&\leq &t_{n}\\\displaystyle {\frac {|\sin(n)|}{n^{3}+n+1}}&\leq &\displaystyle {\frac {1}{n^{3}}}\\|\sin(n)|&\leq &\displaystyle {\frac {n^{3}+n+1}{n^{3}}}\\|\sin(n)|&\leq &\displaystyle {1+{\frac {1}{n^{2}}}+{\frac {1}{n^{3}}}}\end{array}}$
The last inequality is always true since $|\sin(n)|\leq 1$ and the right side is always greater than one. So the series converges.

### Practice Problems with Video Solutions

| 1 | $\sum _{n=1}^{\infty }{\frac {n}{n^{4}+1}}$ | answer convergesconverges | solution | 2 | $\sum _{n=1}^{\infty }{\frac {9^{n}}{3+10^{n}}}$ | answer convergesconverges | solution |