# Real series/Absolutely convergent/Section

## 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 fact 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.

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, 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 ${\displaystyle {}{\frac {\pi ^{2}}{6}}}$.