Real series/Absolutely convergent/Section

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 fact 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.

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