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

Series

We have seen in the last lecture that one can consider a number in the decimal numeral system, meaning an (infinite) sequence of digits between ${\displaystyle {}0}$ and ${\displaystyle {}9}$, as an increasing sequence of rational numbers. For this, the ${\displaystyle {}n}$-th digit after the separator, namely ${\displaystyle {}z_{-n}}$, means ${\displaystyle {}z_{-n}\cdot 10^{-n}}$ and has to be added to the approximation given by the digits before. The sequence of digits describes with the inverse powers of ${\displaystyle {}10}$ 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.

## Definition

Let ${\displaystyle {}{\left(a_{k}\right)}_{k\in \mathbb {N} }}$ be a sequence of real numbers. The series ${\displaystyle {}\sum _{k=0}^{\infty }a_{k}}$ is the sequence ${\displaystyle {}{\left(s_{n}\right)}_{n\in \mathbb {N} }}$ of the partial sums

${\displaystyle {}s_{n}:=\sum _{k=0}^{n}a_{k}\,.}$

If the sequence ${\displaystyle {}{\left(s_{n}\right)}_{n\in \mathbb {N} }}$ converges, then we say that the series converges. In this case, we write also

${\displaystyle \sum _{k=0}^{\infty }a_{k}}$

for its limit,

and this limit is called the sum of the series.

All concepts for sequences carry over to series if we consider a series ${\displaystyle {}\sum _{k=0}^{\infty }a_{k}}$ as the sequence of its partial sums ${\displaystyle {}s_{n}=\sum _{k=0}^{n}a_{k}}$. Like for sequences, it might happen that the sequence does not start with ${\displaystyle {}k=0}$ but later.

## Example

We want to compute the series

${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k(k+1)}}.}$

For this, we give a formula for the ${\displaystyle {}n}$-th partial sum. We have

${\displaystyle {}s_{n}=\sum _{k=1}^{n}{\frac {1}{k(k+1)}}=\sum _{k=1}^{n}{\left({\frac {1}{k}}-{\frac {1}{k+1}}\right)}=1-{\frac {1}{n+1}}={\frac {n}{n+1}}\,.}$

This sequence converges to ${\displaystyle {}1}$, so that the series converges and its sum equals ${\displaystyle {}1}$.

## Lemma

Let

${\displaystyle \sum _{k=0}^{\infty }a_{k}{\text{ and }}\sum _{k=0}^{\infty }b_{k}}$

denote convergent series of real numbers with sums ${\displaystyle {}s}$ and ${\displaystyle {}t}$

respectively. Then the following statements hold.
1. The series ${\displaystyle {}\sum _{k=0}^{\infty }c_{k}}$ given by ${\displaystyle {}c_{k}:=a_{k}+b_{k}}$ is also convergent and its sum is ${\displaystyle {}s+t}$.
2. For ${\displaystyle {}r\in \mathbb {R} }$ also the series ${\displaystyle {}\sum _{k=0}^{\infty }d_{k}}$ given by ${\displaystyle {}d_{k}:=ra_{k}}$ is convergent and its sum is ${\displaystyle {}rs}$.

### Proof

${\displaystyle \Box }$

## Lemma

Let

${\displaystyle \sum _{k=0}^{\infty }a_{k}}$

be a series of real numbers. Then the series is convergent if and only if the following Cauchy-criterion holds: For every ${\displaystyle {}\epsilon >0}$ there exists some ${\displaystyle {}n_{0}}$ such that for all

${\displaystyle {}n\geq m\geq n_{0}\,}$

the estimate

${\displaystyle {}\vert {\sum _{k=m}^{n}a_{k}}\vert \leq \epsilon \,}$

holds.

### Proof

${\displaystyle \Box }$

## Lemma

Let

${\displaystyle \sum _{k=0}^{\infty }a_{k}}$

denote a convergent series of real numbers. Then

${\displaystyle {}\lim _{k\rightarrow \infty }a_{k}=0\,.}$

### Proof

This follows directly from Lemma 9.4 .

${\displaystyle \Box }$

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 harmonic series shows.

## Example

The harmonic series is the series

${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k}}.}$
So this series is about the "infinite sum“ of the unit fractions
${\displaystyle 1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}+{\frac {1}{8}}+\ldots .}$

This series diverges: For the ${\displaystyle {}2^{n}}$ numbers ${\displaystyle {}k=2^{n}+1,\ldots ,2^{n+1}}$, we have

${\displaystyle {}\sum _{k=2^{n}+1}^{2^{n+1}}{\frac {1}{k}}\geq \sum _{k=2^{n}+1}^{2^{n+1}}{\frac {1}{2^{n+1}}}=2^{n}{\frac {1}{2^{n+1}}}={\frac {1}{2}}\,.}$

Therefore,

${\displaystyle {}\sum _{k=1}^{2^{n+1}}{\frac {1}{k}}=1+\sum _{i=0}^{n}\left(\sum _{k=2^{i}+1}^{2^{i+1}}{\frac {1}{k}}\right)\geq 1+(n+1){\frac {1}{2}}\,.}$

Hence, the sequence of the partial sums is unbounded, and so, due to Lemma 9.10 , not convergent.

The following statement is called Leibniz criterion for alternating series.

## Theorem

Let ${\displaystyle {}{\left(x_{k}\right)}_{k\in \mathbb {N} }}$ be an decreasing null sequence of nonnegative real numbers. Then the series ${\displaystyle {}\sum _{k=0}^{\infty }(-1)^{k}x_{k}}$ converges.

### Proof

This proof was not presented in the lecture.
${\displaystyle \Box }$

Absolutely convergent series

## Definition

${\displaystyle \sum _{k=0}^{\infty }a_{k}}$

of real numbers is called absolutely convergent, if the series

${\displaystyle \sum _{k=0}^{\infty }\vert {a_{k}}\vert }$
converges.

## Lemma

### Proof

Let ${\displaystyle {}\epsilon >0}$ be given. We use the Cauchy-criterion. Since the series converges absolutely, there exists some ${\displaystyle {}n_{0}}$ such that for all ${\displaystyle {}n\geq m\geq n_{0}}$ the estimate

${\displaystyle {}\vert {\sum _{k=m}^{n}\vert {a_{k}}\vert \,}\vert =\sum _{k=m}^{n}\vert {a_{k}}\vert \leq \epsilon \,}$

holds. Therefore,

${\displaystyle {}\vert {\sum _{k=m}^{n}a_{k}}\vert \leq \vert {\sum _{k=m}^{n}\vert {a_{k}}\vert \,}\vert \leq \epsilon \,,}$

which means the convergence.

${\displaystyle \Box }$

## Example

A convergent series does not in general converge absolutely, the converse of Lemma 9.9 does not hold. Due to the Leibniz criterion, the alternating harmonic series

${\displaystyle {}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}=1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+{\frac {1}{5}}-\ldots \,}$

converges, and its sum is ${\displaystyle {}\ln 2}$, 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 direct comparison test.

## Lemma

Let ${\displaystyle {}\sum _{k=0}^{\infty }b_{k}}$ be a convergent series of real numbers and ${\displaystyle {}{\left(a_{k}\right)}_{k\in \mathbb {N} }}$ a sequence of real numbers fulfilling ${\displaystyle {}\vert {a_{k}}\vert \leq b_{k}}$ for all ${\displaystyle {}k}$. Then the series

${\displaystyle \sum _{k=0}^{\infty }a_{k}}$

### Proof

This follows directly from the Cauchy-criterion.

${\displaystyle \Box }$

## Example

We want to determine whether the series

${\displaystyle {}\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}=1+{\frac {1}{4}}+{\frac {1}{9}}+{\frac {1}{16}}+{\frac {1}{25}}+\ldots \,}$

converges. We use the direct comparison test and Example 9.2 , where we have shown the convergence of ${\displaystyle {}\sum _{k=1}^{n}{\frac {1}{k(k+1)}}}$. For ${\displaystyle {}k\geq 2}$ we have

${\displaystyle {}{\frac {1}{k^{2}}}\leq {\frac {1}{k(k-1)}}\,.}$

Hence, ${\displaystyle {}\sum _{k=2}^{\infty }{\frac {1}{k^{2}}}}$ converges and therefore also ${\displaystyle {}\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}}$. This does not say much about the exact value of the sum. With much more advanced methods, one can show that this sum equals ${\displaystyle {}{\frac {\pi ^{2}}{6}}}$.

Geometric series and ratio test

The series ${\displaystyle {}\sum _{k=0}^{\infty }x^{k}}$ is called geometric series for ${\displaystyle {}x\in \mathbb {R} }$, so this is the sum

${\displaystyle 1+x+x^{2}+x^{3}+\ldots .}$

The convergence depends heavily on the modulus of ${\displaystyle {}x}$.

## Theorem

For all real numbers ${\displaystyle {}x}$ with ${\displaystyle {}\vert {x}\vert <1}$, the geometric series ${\displaystyle {}\sum _{k=0}^{\infty }x^{k}}$ converges absolutely, and the sum equals

${\displaystyle {}\sum _{k=0}^{\infty }x^{k}={\frac {1}{1-x}}\,.}$

### Proof

For every ${\displaystyle {}x}$ and every ${\displaystyle {}n\in \mathbb {N} }$ we have the relation

${\displaystyle {}(x-1){\left(\sum _{k=0}^{n}x^{k}\right)}=x^{n+1}-1\,}$

and hence for the partial sums the relation (for ${\displaystyle {}x\neq 1}$ )

${\displaystyle {}s_{n}=\sum _{k=0}^{n}x^{k}={\frac {x^{n+1}-1}{x-1}}\,}$

holds. For ${\displaystyle {}n\rightarrow \infty }$ and ${\displaystyle {}\vert {x}\vert <1}$ this converges to ${\displaystyle {}{\frac {-1}{x-1}}={\frac {1}{1-x}}}$ because of Lemma 8.1 and exercise *****.

${\displaystyle \Box }$

The following statement is called ratio test.

## Theorem

Let

${\displaystyle \sum _{k=0}^{\infty }a_{k}}$

be a series of real numbers. Suppose there exists a real number ${\displaystyle {}q}$ with ${\displaystyle {}0\leq q<1}$, and a ${\displaystyle {}k_{0}}$ with

${\displaystyle {}\vert {\frac {a_{k+1}}{a_{k}}}\vert \leq q\,}$

for all ${\displaystyle {}k\geq k_{0}}$ (in particular ${\displaystyle {}a_{k}\neq 0}$ for ${\displaystyle {}k\geq k_{0}}$). Then the series ${\displaystyle {}\sum _{k=0}^{\infty }a_{k}}$ converges absolutely.

### Proof

The convergence does not change (though the sum) when we change finitely many members of the series. Therefore, we can assume ${\displaystyle {}k_{0}=0}$. Moreover, we can assume that all ${\displaystyle {}a_{k}}$ are positive real numbers. Then

${\displaystyle {}a_{k}={\frac {a_{k}}{a_{k-1}}}\cdot {\frac {a_{k-1}}{a_{k-2}}}\cdots {\frac {a_{1}}{a_{0}}}\cdot a_{0}\leq a_{0}\cdot q^{k}\,.}$

Hence, the convergence follows from the comparison test and the convergence of the geometric series.

${\displaystyle \Box }$

## Example

The Koch snowflakes are given by the sequence of plane geometric shapes ${\displaystyle {}K_{n}}$, which are defined recursively in the following way: The starting object ${\displaystyle {}K_{0}}$ is an equilateral triangle. The object ${\displaystyle {}K_{n+1}}$ is obtained from ${\displaystyle {}K_{n}}$ by replacing in each edge of ${\displaystyle {}K_{n}}$ the third in the middle by the corresponding equilateral triangle showing outside.

Let ${\displaystyle {}A_{n}}$ denote the area and ${\displaystyle {}L_{n}}$ the length of the boundary of the ${\displaystyle {}n}$-th Koch snowflake. We want to show that the sequence ${\displaystyle {}A_{n}}$ converges and that the sequence ${\displaystyle {}L_{n}}$ diverges to ${\displaystyle {}\infty }$.

The number of edges of ${\displaystyle {}K_{n}}$ is ${\displaystyle {}3\cdot 4^{n}}$, since in each division step, one edge is replaced by four edges. Their length is ${\displaystyle {}1/3}$ of the length of a previous edge. Let ${\displaystyle {}r}$ denote the base length of the starting equilateral triangle. Then ${\displaystyle {}K_{n}}$ consists of ${\displaystyle {}3\cdot 4^{n}}$ edges of length ${\displaystyle {}r{\left({\frac {1}{3}}\right)}^{n}}$ and the length of all edges of ${\displaystyle {}K_{n}}$ together is

${\displaystyle {}L_{n}=3\cdot 4^{n}r{\left({\frac {1}{3}}\right)}^{n}=3r{\left({\frac {4}{3}}\right)}^{n}\,.}$

Because of ${\displaystyle {}{\frac {4}{3}}>1}$, this diverges to ${\displaystyle {}\infty }$.

When we turn from ${\displaystyle {}K_{n}}$ to ${\displaystyle {}K_{n+1}}$, 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 ${\displaystyle {}s}$ is ${\displaystyle {}{\frac {\sqrt {3}}{4}}s^{2}}$. So in the step from ${\displaystyle {}K_{n}}$ to ${\displaystyle {}K_{n+1}}$ there are ${\displaystyle {}3\cdot 4^{n}}$ triangles added with area ${\displaystyle {}{\frac {\sqrt {3}}{4}}{\left({\frac {1}{3}}\right)}^{2(n+1)}r^{2}={\frac {\sqrt {3}}{4}}r^{2}{\left({\frac {1}{9}}\right)}^{n+1}}$. The total area of ${\displaystyle {}K_{n}}$ is therefore

{\displaystyle {}{\begin{aligned}&\,{\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)}.\end{aligned}}}

If we forget the ${\displaystyle {}1}$ and the factor ${\displaystyle {}{\frac {3}{4}}}$, which does not change the convergence property, we get in the bracket a partial sum of the geometric series for ${\displaystyle {}{\frac {4}{9}}}$, and this converges.