# Real series/Introduction/Section

## 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 fact.

${\displaystyle \Box }$

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 fact, 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.
${\displaystyle \Box }$