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 ${}0$ and ${}9$, as an increasing sequence of rational numbers. For this, the ${}n$-th digit after the separator, namely ${}z_{-n}$, means ${}z_{-n}\cdot 10^{-n}$ 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.

Definition

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

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

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

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

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 ${}\sum _{k=0}^{\infty }a_{k}$ as the sequence of its partial sums ${}s_{n}=\sum _{k=0}^{n}a_{k}$. Like for sequences, it might happen that the sequence does not start with ${}k=0$ but later.

Example

We want to compute the series

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

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

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

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

Lemma

Let

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

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

respectively. Then the following statements hold.

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

Proof

See Exercise 9.8 .

$\Box$

Lemma

Let

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

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

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

the estimate

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

holds.

Proof

See Exercise 9.9 .

$\Box$

Lemma

Let

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

denote a convergent series of real numbers. Then

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

Proof

This follows directly from Lemma 9.4 .

$\Box$

Nikolaus of Oresme (1330-1382) proved that the harmonic series diverges.

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

$\sum _{k=1}^{\infty }{\frac {1}{k}}.$

So this series is about the "infinite sum“ of the unit fractions

$1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}+{\frac {1}{8}}+\ldots .$

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

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

Therefore,

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

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

The divergence of the harmonic series implies that one can construct with equal building bricks an arbitrary large overhang.

The following statement is called Leibniz criterion for alternating series.

Theorem

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

Proof

This proof was not presented in the lecture.

$\Box$

Absolutely convergent series

Definition

A series

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

of real numbers is called absolutely convergent, if the series

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

converges.

Lemma

An absolutely convergent series of real numbers converges.

Proof

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

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

holds. Therefore,

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

which means the convergence.

$\Box$

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

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

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 direct comparison test.

Lemma

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

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

is absolutely convergent.

Proof

This follows directly from the Cauchy-criterion.

$\Box$

Example

We want to determine whether the series

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

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

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

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

Geometric series and ratio test

This image illustrates the behavior of the geometric series for ${}x={\frac {1}{4}}$. Suppose that the length of the square is ${}2$. Then we can put the geometric series threetimes inside this square. The area of the three series is ${}{\frac {4}{3}}$.

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

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

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

Theorem

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

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

Proof

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

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

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

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

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

$\Box$

The following statement is called ratio test.

Theorem

Let

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

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

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

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

Proof

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

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

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

$\Box$

Example

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

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

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

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

${}{\begin{aligned}&\,{\frac {\sqrt {3}}{4}}r^{2}{\left(1+3{\frac {1}{9}}+12{\left({\frac {1}{9}}\right)}^{2}+48{\left({\frac {1}{9}}\right)}^{3}+\cdots +3\cdot 4^{n-1}{\left({\frac {1}{9}}\right)}^{n}\right)}\\&={\frac {\sqrt {3}}{4}}r^{2}{\left(1+{\frac {3}{4}}{\left({\frac {4}{9}}\right)}^{1}+{\frac {3}{4}}{\left({\frac {4}{9}}\right)}^{2}+{\frac {3}{4}}{\left({\frac {4}{9}}\right)}^{3}+\cdots +{\frac {3}{4}}{\left({\frac {4}{9}}\right)}^{n}\right)}.\end{aligned}}$

If we forget the ${}1$ and the factor ${}{\frac {3}{4}}$, 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.

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