# Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 9/latex

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\zwischenueberschrift{Series}
We have seen in the last lecture that one can consider a number in the decimal numeral system, meaning an \zusatzklammer {infinite} sequence of digits between \mathkor {0} {and} {9} {,} as an increasing sequence of rational numbers. For this, the $n$-th digit after the separator, namely \mathl{z_{-n}}{,} means \mathl{z_{-n} \cdot 10^{-n}} and has to be added to the approximation given by the digits before. The sequence of digits describes with the inverse powers of $10$ the difference between the approximating sequence, and the members in the approximating sequence are gained by summing up these differences. This viewpoint leads to the concept of a series.
\inputdefinition { } {
Let \mathl{{ \left( a_k \right) }_{k \in \N }} be a sequence of real numbers. The \definitionswort {series} \mathl{\sum_{ k = 0}^\infty a_{ k }} is the sequence \mathl{{ \left( s_n \right) }_{n \in \N }} of the \definitionswort {partial sums} \mavergleichskettedisp {\vergleichskette { s_n } { \defeq} { \sum_{ k = 0}^n a_{ k } } { } { } { } { } { } { } }{.} If the sequence \mathl{{ \left( s_n \right) }_{n \in \N }} converges, then we say that the \definitionswort {series converges}{.} In this case, we write also \mathdisp {\sum_{ k = 0}^\infty a_{ k }} { } for its limit,
All concepts for sequences carry over to series if we consider a series \mathl{\sum_{ k = 0}^\infty a_{ k }} as the sequence of its partial sums \mavergleichskette {\vergleichskette { s_n } { = }{ \sum_{ k = 0}^n a_{ k } } { }{ } { }{ } { }{ } }{.} Like for sequences, it might happen that the sequence does not start with \mavergleichskette {\vergleichskette {k } { = }{ 0 } { }{ } { }{ } { }{ } } but later.
\inputbeispiel {
We want to compute the series \mathdisp {\sum_{k=1}^\infty { \frac{ 1 }{ k(k+1) } }} { . } For this, we give a formula for the $n$-th partial sum. We have \mavergleichskettedisp {\vergleichskette { s_n } { =} { \sum_{k = 1}^n { \frac{ 1 }{ k(k+1) } } } { =} { \sum_{k = 1}^n { \left( { \frac{ 1 }{ k } } - { \frac{ 1 }{ k+1 } } \right) }
} { =} { 1- { \frac{ 1 }{ n+1 } } } { =} { { \frac{ n }{ n+1 } } } }{.} This sequence converges to $1$, so that the series converges and its sum equals $1$.
\faktsituation {Let \mathdisp {\sum_{ k = 0}^\infty a_{ k } \text{ and } \sum_{ k = 0}^\infty b_{ k }} { } denote convergent series of real numbers with sums \mathkor {s} {and} {t} respectively.} \faktuebergang {Then the following statements hold.} \faktfolgerung {\aufzaehlungzwei {The series \mathl{\sum_{ k = 0}^\infty c_{ k }} given by \mavergleichskette {\vergleichskette { c_k } { \defeq }{ a_k+b_k } { }{ } { }{ } { }{ } } is also convergent and its sum is \mathl{s+t}{.} } {For \mavergleichskette {\vergleichskette { r } { \in }{ \R } { }{ } { }{ } { }{ } } also the series \mathl{\sum_{ k = 0}^\infty d_{ k }} given by \mavergleichskette {\vergleichskette { d_k } { \defeq }{ r a_k } { }{ } { }{ } { }{ } } is convergent and its sum is $r s$. }} \faktzusatz
\faktsituation {Let \mathdisp {\sum_{ k = 0}^\infty a_{ k }} { } be a series of real numbers.} \faktfolgerung {Then the series is convergent if and only if the following \stichwort {Cauchy-criterion} holds: For every \mavergleichskette {\vergleichskette {\epsilon } { > }{0 } { }{ } { }{ } { }{ } } there exists some $n_0$ such that for all \mavergleichskettedisp {\vergleichskette {n } { \geq} {m } { \geq} {n_0 } { } { } { } { } } the estimate \mavergleichskettedisp {\vergleichskette { \betrag { \sum_{k = m}^n a_k } } { \leq} { \epsilon } { } { } { } { } { } { } } holds.} \faktzusatz
\faktsituation {Let \mathdisp {\sum_{ k = 0}^\infty a_{ k }} { } denote a convergent series of real numbers.} \faktfolgerung {Then \mavergleichskettedisp {\vergleichskette { \lim_{ k \rightarrow \infty} a_{ k } } { =} { 0 } { } { } { } { } { } { } }{.}} \faktzusatz } {
This follows directly from Lemma 9.4 .
It is therefore a necessary condition for the convergence of a series that its members form a null sequence. This condition is not sufficient, as the \stichwort {harmonic series} shows. \inputbeispiel {
The \definitionswort {harmonic series} is the series
\mathdisp {1 + { \frac{ 1 }{ 2 } } + { \frac{ 1 }{ 3 } } + { \frac{ 1 }{ 4 } } + { \frac{ 1 }{ 5 } } + { \frac{ 1 }{ 6 } } + { \frac{ 1 }{ 7 } } + { \frac{ 1 }{ 8 } } + \ldots} { . } This series diverges: For the $2^{n}$ numbers \mavergleichskette {\vergleichskette { k } { = }{ 2^n +1 , \ldots , 2^{n+1} } { }{ } { }{ } { }{ } }{,} we have \mavergleichskettedisp {\vergleichskette { \sum_{k = 2^n+1}^{ 2^{n+1} } \frac{1}{k} } { \geq} { \sum_{k = 2^n+1}^{ 2^{n+1} } \frac{1}{2^{n+1} } } { =} { 2^n \frac{1}{2^{n+1} } } { =} { \frac{1}{2} } { } { } }{.} Therefore, \mavergleichskettedisp {\vergleichskette { \sum_{k = 1}^{ 2^{n+1} } \frac{1}{k} } { =} {1+ \sum_{i = 0}^n \left( \sum_{k = 2^{i} +1 }^{ 2^{i+1} } \frac{1}{k} \right)} { } { } { \geq} {1 + (n+1) \frac{1}{2} } { } { } }{.} Hence, the sequence of the partial sums is unbounded, and so, due to Lemma 9.10 , not convergent.
The following statement is called \stichwort {Leibniz criterion for alternating series} {.}
\faktsituation {Let \mathl{{ \left( x_k \right) }_{k \in \N }} be an decreasing null sequence of nonnegative real numbers.} \faktfolgerung {Then the series \mathl{\sum_{ k= 0}^\infty (-1)^k x_k} converges.} \faktzusatz } {Series/Real numbers/Leibniz criterion/Fact/Proof
\zwischenueberschrift{Absolutely convergent series}
\inputdefinition { } {
A series \mathdisp {\sum_{ k = 0}^\infty a_{ k }} { } of real numbers is called \definitionswort {absolutely convergent}{,} if the series \mathdisp {\sum_{ k = 0}^\infty \betrag { a_k }} { }
\faktsituation \faktvoraussetzung {An absolutely convergent series of real numbers} \faktfolgerung {converges.} \faktzusatz } {
Let \mavergleichskette {\vergleichskette { \epsilon } { > }{ 0 } { }{ } { }{ } { }{ } } be given. We use the Cauchy-criterion. Since the series converges absolutely, there exists some $n_0$ such that for all \mavergleichskette {\vergleichskette {n } { \geq }{ m } { \geq }{ n_0 } { }{ } { }{ } } the estimate \mavergleichskettedisp {\vergleichskette { \betrag { \sum_{k = m}^n \betrag { a_k } \, } } { =} { \sum_{k = m}^n \betrag { a_k } } { \leq} { \epsilon } { } { } { } { } } holds. Therefore, \mavergleichskettedisp {\vergleichskette { \betrag { \sum_{k = m}^n a_k } } { \leq} { \betrag { \sum_{k = m}^n \betrag { a_k } \, } } { \leq} { \epsilon } { } { } { } { } }{,} which means the convergence.
\inputbeispiel {
A convergent series does not in general converge absolutely, the converse of Lemma 9.9
does not hold. Due to the Leibniz criterion, the \stichwort {alternating harmonic series} \mavergleichskettedisp {\vergleichskette { \sum_{n = 1}^\infty \frac{ (-1)^{n+1} }{n} } { =} { 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \ldots } { } { } { } { } { } { } } converges, and its sum is $\ln 2$, a result we can not prove here. However, the corresponding absolute series is just the harmonic series, which diverges due to Example 9.6 .
The following statement is called the \stichwort {direct comparison test} {.}
\faktsituation \faktvoraussetzung {Let \mathl{\sum_{ k = 0}^\infty b_{ k }} be a convergent series of real numbers and \mathl{{ \left( a_k \right) }_{k \in \N }} a sequence of real numbers fulfilling \mavergleichskette {\vergleichskette { \betrag { a_k } } { \leq }{ b_k } { }{ } { }{ } { }{ } } for all $k$.} \faktfolgerung {Then the series \mathdisp {\sum_{ k = 0}^\infty a_{ k }} { } is absolutely convergent.} \faktzusatz } {
This follows directly from the Cauchy-criterion.
\inputbeispiel {
We want to determine whether the series \mavergleichskettedisp {\vergleichskette { \sum_{k = 1}^\infty { \frac{ 1 }{ k^2 } } } { =} { 1 + { \frac{ 1 }{ 4 } } + { \frac{ 1 }{ 9 } } + { \frac{ 1 }{ 16 } } + { \frac{ 1 }{ 25 } } + \ldots } { } { } { } { } { } { } } converges. We use the direct comparison test and Example 9.2 , where we have shown the convergence of \mathl{\sum_{k = 1}^n { \frac{ 1 }{ k(k+1) } }}{.} For \mavergleichskette {\vergleichskette {k } { \geq }{2 } { }{ } { }{ } { }{ } } we have \mavergleichskettedisp {\vergleichskette { { \frac{ 1 }{ k^2 } } } { \leq} { { \frac{ 1 }{ k(k-1) } } } { } { } { } { } { } { } }{.} Hence, \mathl{\sum_{k=2}^\infty { \frac{ 1 }{ k^2 } }} converges and therefore also \mathl{\sum_{k=1}^\infty { \frac{ 1 }{ k^2 } }}{.} This does not say much about the exact value of the sum. With much more advanced methods, one can show that this sum equals \mathl{{ \frac{ \pi^2 }{ 6 } }}{.}
\zwischenueberschrift{Geometric series and ratio test}
The series \mathl{\sum_{ k = 0}^\infty x^k} is called \stichwort {geometric series} for \mavergleichskette {\vergleichskette {x } { \in }{ \R } { }{ } { }{ } { }{ } }{,} so this is the sum \mathdisp {1+x+x^2+x^3+ \ldots} { . } The convergence depends heavily on the modulus of $x$.
For all real numbers $x$ with \mavergleichskette {\vergleichskette { \betrag { x } } { < }{1 } { }{ } { }{ } { }{ } }{,} the geometric series \mathl{\sum^\infty_{k = 0} x^k} converges absolutely, and the sum equals \mavergleichskettedisp {\vergleichskette { \sum^\infty_{k = 0} x^k } { =} { { \frac{ 1 }{ 1-x } } } { } { } { } { } { } { } }{.} } {
For every $x$ and every \mavergleichskette {\vergleichskette {n } { \in }{ \N } { }{ } { }{ } { }{ } } we have the relation \mavergleichskettedisp {\vergleichskette { (x-1) { \left(\sum_{ k = 0}^n x^k\right) } } { =} { x^{n+1} -1 } { } { } { } { } { } { } } and hence for the partial sums the relation (for \mavergleichskette {\vergleichskette { x } { \neq }{ 1 } { }{ } { }{ } { }{ } } ) \mavergleichskettedisp {\vergleichskette { s_n } { =} { \sum_{ k = 0}^n x^k } { =} { \frac{ x^{n+1} -1}{ x-1 } } { } { } { } } holds. For \mathl{n \rightarrow \infty} and \mavergleichskette {\vergleichskette { \betrag { x } } { < }{ 1 } { }{ } { }{ } { }{ } } this converges to \mavergleichskette {\vergleichskette { \frac{-1}{x-1} } { = }{ \frac{1}{1-x} } { }{ } { }{ } { }{ } } because of Lemma 8.1
and exercise *****.
The following statement is called \stichwort {ratio test} {.}
\faktsituation {Let \mathdisp {\sum_{ k = 0}^\infty a_{ k }} { } be a series of real numbers.} \faktvoraussetzung {Suppose there exists a real number $q$ with \mavergleichskette {\vergleichskette { 0 } { \leq }{ q } { < }{ 1 } { }{ } { }{ } }{,} and a $k_0$ with \mavergleichskettedisp {\vergleichskette { \betrag { { \frac{ a_{k+1} }{ a_k } } } } { \leq} { q } { } { } { } { } { } { } } for all \mavergleichskette {\vergleichskette { k } { \geq }{ k_0 } { }{ } { }{ } { }{ } } \zusatzklammer {in particular \mavergleichskette {\vergleichskette { a_k } { \neq }{0 } { }{ } { }{ } { }{ } } for \mavergleichskette {\vergleichskette { k } { \geq }{ k_0 } { }{ } { }{ } { }{ } }} {.}} \faktfolgerung {Then the series \mathl{\sum_{ k = 0}^\infty a_{ k }} converges absolutely.} \faktzusatz } {
The convergence does not change \zusatzklammer {though the sum} when we change finitely many members of the series. Therefore, we can assume \mavergleichskette {\vergleichskette { k_0 } { = }{ 0 } { }{ } { }{ } { }{ } }{.} Moreover, we can assume that all $a_k$ are positive real numbers. Then \mavergleichskettedisp {\vergleichskette { a_k } { =} { \frac{a_k}{a_{k-1} } \cdot \frac{a_{k-1} }{a_{k-2} } \cdots \frac{a_1}{a_{0} } \cdot a_0 } { \leq} { a_0 \cdot q^k } { } { } { } { } }{.} Hence, the convergence follows from the comparison test and the convergence of the geometric series.
\inputbeispiel {
The \stichwort {Koch snowflakes} are given by the sequence of plane geometric shapes $K_n$, which are defined recursively in the following way: The starting object $K_0$ is an equilateral triangle. The object $K_{n+1}$ is obtained from $K_n$ by replacing in each edge of $K_n$ the third in the middle by the corresponding equilateral triangle showing outside.
Let $A_n$ denote the area and $L_n$ the length of the boundary of the $n$-th Koch snowflake. We want to show that the sequence $A_n$ converges and that the sequence $L_n$ diverges to $\infty$.
The number of edges of $K_n$ is \mathl{3 \cdot 4^n}{,} since in each division step, one edge is replaced by four edges. Their length is \mathl{1/3} of the length of a previous edge. Let $r$ denote the base length of the starting equilateral triangle. Then $K_n$ consists of \mathl{3 \cdot 4^n} edges of length \mathl{r { \left( { \frac{ 1 }{ 3 } } \right) }^n} and the length of all edges of $K_n$ together is \mavergleichskettedisp {\vergleichskette { L_n } { =} { 3 \cdot 4^n r { \left( { \frac{ 1 }{ 3 } } \right) }^n } { =} { 3 r { \left( { \frac{ 4 }{ 3 } } \right) }^n } { } { } { } { } }{.} Because of \mavergleichskette {\vergleichskette { { \frac{ 4 }{ 3 } } } { > }{ 1 } { }{ } { }{ } { }{ } }{,} this diverges to $\infty$.
When we turn from $K_{n}$ to \mathl{K_{n+1}}{,} there will be for every edge a new triangle whose side length is a third of the edge length. The area of an equilateral triangle with side length $s$ is \mathl{{ \frac{ \sqrt{3} }{ 4 } } s^2}{.} So in the step from \mathl{K_{n}} to \mathl{K_{n+1}} there are \mathl{3 \cdot 4^{n}} triangles added with area \mavergleichskette {\vergleichskette { { \frac{ \sqrt{3} }{ 4 } } { \left( { \frac{ 1 }{ 3 } } \right) }^{2(n+1)} r^2 } { = }{ { \frac{ \sqrt{3} }{ 4 } } r^2 { \left( { \frac{ 1 }{ 9 } } \right) }^{n+1} } { }{ } { }{ } { }{ } }{.} The total area of $K_n$ is therefore \mavergleichskettealign {\vergleichskettealign { } { \,} { { \frac{ \sqrt{3} }{ 4 } } r^2 { \left( 1 + 3 { \frac{ 1 }{ 9 } } + 12 { \left( { \frac{ 1 }{ 9 } } \right) } ^2 + 48 { \left( { \frac{ 1 }{ 9 } } \right) }^3 + \cdots + 3\cdot 4^{n-1} { \left( { \frac{ 1 }{ 9 } } \right) }^{n} \right) } } { =} { { \frac{ \sqrt{3} }{ 4 } } r^2 { \left( 1 + { \frac{ 3 }{ 4 } } { \left( { \frac{ 4 }{ 9 } } \right) }^1 + { \frac{ 3 }{ 4 } } { \left( { \frac{ 4 }{ 9 } } \right) }^2 + { \frac{ 3 }{ 4 } } { \left( { \frac{ 4 }{ 9 } } \right) }^3 + \cdots + { \frac{ 3 }{ 4 } } { \left( { \frac{ 4 }{ 9 } } \right) }^{n} \right) } } { } { } { } { } }{.} If we forget the $1$ and the factor \mathl{{ \frac{ 3 }{ 4 } }}{,} which does not change the convergence property, we get in the bracket a partial sum of the geometric series for ${ \frac{ 4 }{ 9 } }$, and this converges.