# Kurs:Analysis 1 (TU Dortmund)/§4 Unendliche Reihen

Zur Navigation springen Zur Suche springen

## 4.1 Unendliche Reihen

Setze ${\displaystyle S_{n}:=\sum _{k=0}^{n}a_{k}}$. Wenn ${\displaystyle \lim _{n\to \infty }s_{n}}$ existiert, so sehen wir (*) ${\displaystyle \sum _{k=0}^{\infty }a_{k}=\lim _{n\to \infty }S_{n}}$.

${\displaystyle \sum _{k=0}^{\infty }a_{k}}$ heißt unendliche Reihe, ${\displaystyle S_{n}}$ ihre n-te Partialsumme.

${\displaystyle \sum _{k=0}^{\infty }a_{k}}$ heißt konvergent, wenn (*) existiert, sonst divergent.

## 4.2 Notwendige Bedingungen für die Konvergenz:

(i) ${\displaystyle a_{k}\to 0(k\to \infty )}$
(ii) ${\displaystyle r_{n}=\sum _{k=n+1}^{\infty }a_{k}\to 0(n\to \infty )}$ (Reihenrest)

Beweis: ${\displaystyle S_{n}=\sum _{k=0}^{n}a_{k}\to s(n\to \infty )}$;

${\displaystyle a_{n}=S_{n}-S_{n-1}\to S-S=0(n\to \infty )}$

${\displaystyle r_{n}=s-S_{n}\to s-s=0(n\to \infty )}$

## 4.3 Teleskopreihen

${\displaystyle (a_{k})}$ sei Nullfolge. Dann konvergiert ${\displaystyle \sum _{k=0}^{\infty }(a_{k}-a_{k+1})=a_{0}}$

dann ${\displaystyle \sum _{k=0}^{\infty }(a_{k}-a_{k+p})=a_{0}+a_{1}+...+a_{p-1}}$.

Beweis: ${\displaystyle n\geq pS_{n}=\sum _{k=0}^{n}(a_{k}-a_{k+p})=\sum _{k=0}^{n}a_{k}-\sum _{k=0}^{n}a_{k+p}=\sum _{k=0}^{n}a_{k}-\sum _{j=p}^{n+p}a_{j}}$

...

Beispiel: ${\displaystyle \sum _{n=0}^{\infty }{\frac {1}{n^{2}+4n+3}}}$

${\displaystyle n^{2}+4n+3=(n+1)(n+3)\Rightarrow {\frac {1}{(n+1)(n+3)}}={\frac {\frac {1}{2}}{(n+1)}}-{\frac {\frac {1}{2}}{(n+3)}}}$

${\displaystyle \sum _{n=0}^{\infty }{\frac {1}{n^{2}+4n+3}}=\sum _{n=0}^{\infty }{\frac {1}{2}}({\frac {1}{n+1}}-{\frac {1}{n+3}})=a_{0}+a_{1}={\frac {\frac {1}{2}}{0+1}}+{\frac {\frac {1}{2}}{1+3}}={\frac {1}{2}}+{\frac {1}{8}}={\frac {5}{8}}}$.

Achtung! ${\displaystyle \sum _{n=0}^{\infty }\left({\frac {\frac {1}{2}}{(n+1)}}-{\frac {\frac {1}{2}}{(n+3)}}\right)\neq \sum _{n=0}^{\infty }{\frac {\frac {1}{2}}{(n+1)}}-\sum _{n=0}^{\infty }{\frac {\frac {1}{2}}{(n+3)}}}$, da die einzelnen Summen divergent sind.

...

## 4.4 Satz

Wenn ${\displaystyle \sum _{k=0}^{\infty }a_{k}}$ und ${\displaystyle \sum _{k=0}^{\infty }b_{k}}$ konvergent, dann gilt

${\displaystyle \sum _{k=0}^{\infty }(a_{k}+b_{k})=\sum _{k=0}^{\infty }a_{k}+\sum _{k=0}^{\infty }b_{k}}$

und

${\displaystyle \sum _{k=0}^{\infty }ca_{k}=c\sum _{k=0}^{\infty }a_{k}(c\in \mathbb {R} }$ )

Beweis, siehe Folgen.

## 4.5 Cauchykriterium (für Reihen)

Die Reihe ${\displaystyle \sum _{k=0}^{\infty }a_{k}}$ ist genau dann konvergent, wenn es zu jedem ${\displaystyle \varepsilon >0}$ ein ${\displaystyle n_{0}}$ gibt und ${\displaystyle \left|\sum _{k=m+1}^{n}a_{k}\right|<\varepsilon }$ für alle ${\displaystyle n>m\geq n_{0}}$.

Beweis: ...

## 4.6 Die harmonische Reihe

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n}}}$ ist divergent.

Beweis: ...

## 4.7 Leibnizkriterium (alternierende Reihen)

Sei ${\displaystyle (a_{k})_{k\geq 0}}$ eine monoton fallende Nullfolge.

Dann konvergiert ${\displaystyle \sum _{k=0}^{\infty }(-1)^{k}a_{k}}$ mit

${\displaystyle S_{2n-1}\leq s\leq S_{2n}\ \ \ \ (S_{n}=\sum _{k=0}^{\infty }(-1)^{k}a_{k})}$

Beweis:

${\displaystyle S_{2n+2}-S_{2n}=a_{2n-2}-a_{2n+1}\leq 0\ \ \ S_{2n}\downarrow }$
${\displaystyle S_{2n+3}-S_{2n+1}=-a_{2n+3}+a_{2n+2}\geq 0\ \ \ S_{2n+1}\uparrow }$
${\displaystyle S_{2n}-S_{2n-1}=a_{2n}\geq 0}$
${\displaystyle S_{1}\leq S_{3}\leq ...\leq S_{2n-1}\leq s\leq S_{2n}\leq S_{2n-2}\leq ...\leq S_{0}}$
${\displaystyle \lim _{n\to \infty }S_{2n-1}=\lim _{n\to \infty }S_{2n}=s}$ da ${\displaystyle a_{2n}\rightarrow 0(n\rightarrow \infty )}$
Siehe auch
Leibniz-Kriterium

## 4.8 Beispiel - Die alternierende harmonische Reihe

Die Reihe ${\displaystyle \sum _{n=1}^{\infty }(-1)^{n-1}{\frac {1}{n}}=\sum _{k=0}^{\infty }(-1)^{k}{\frac {1}{k+1}}}$ konvergiert, da ${\displaystyle (a_{k}={\frac {1}{k+1}}\downarrow nullfolge)}$

Wert s: ${\displaystyle {\frac {1}{2}}\leq s\leq 1}$

${\displaystyle 1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}\leq s\leq 1-{\frac {1}{2}}+{\frac {1}{3}}}$

## 4.9 Abel-Dirichletkriterium

Sei ${\displaystyle (a_{k})}$ eine monoton fallende Nullfolge und sei ${\displaystyle (b_{k})}$ eine Folge mit: ${\displaystyle B_{k}=\sum _{k=0}^{n}b_{k}}$ sei beschränkt. Dann konvergiert ${\displaystyle \sum _{k=0}^{\infty }a_{k}b_{k}=\sum _{k=0}^{\infty }B_{k}(a_{k+1}-a_{k})}$

Bemerkung:

${\displaystyle b_{k}=(-1)^{k},\ \ B_{k}={\begin{cases}1\ \ \ k\ {\text{gerade}}\\0\ \ \ k\ {\text{ungerade}}\end{cases}}}$

Beweis: "abelsche partielle Summation"

${\displaystyle {\begin{matrix}\sum _{k=m-1}^{n}a_{k}b_{k}&=&\sum _{k=m+1}^{n}a_{k}(B_{k}-B_{k-1})\\&=&\sum _{k=m+1}^{n}a_{k}B_{k}-\sum _{j=m}^{n-1}a_{j+1}B_{j}&k-1=j\\&=&a_{n}B_{n}-a_{m+1}B_{m}+\sum _{k=m+1}^{n-1}a_{k}B_{k}-\sum _{k=m+1}^{n-1}a_{k+1}B_{k}\\&=&a_{n}B_{n}-a_{m+1}B_{m}+\sum _{k=m+1}^{n-1}(a_{k}-a_{k+1})B_{k}\end{matrix}}}$

Konvergenz:

${\displaystyle \left|B_{k}\right|\leq B(k\in \mathbb {N} _{0})}$
${\displaystyle \left|\sum _{k=m+1}^{n}a_{k}b_{k}\right|\leq \underbrace {a_{n}B} _{\leq 0}+\underbrace {a_{m+1}} _{\leq 0}+\sum _{k=m+1}^{n-1}\underbrace {a_{n}-a_{n+1}} _{\geq 0}B=B(a_{n}+a_{m+1}+a_{m+a}-a_{n})=2B_{am+1}}$
zu ${\displaystyle \epsilon >0\ \exists n_{0}:a_{m+1}\leq \epsilon /B(m\geq n_{0})}$
d.h. ${\displaystyle \left|\sum _{k=m+1}^{n}a_{k}b_{k}\right|<\epsilon /B(n>m\geq n_{0})}$ d.h. CK erfüllt

Reihenwert

${\displaystyle (m=-1,B_{-1}=0)}$
${\displaystyle \sum _{k=0}^{n}a_{k}b_{k}=a_{n}B_{n}-a_{0}B_{-1}+\sum _{k=0}^{n-1}(a_{k}-a_{k+1})B_{k}}$
${\displaystyle n\rightarrow \infty :\sum _{k=0}^{\infty }a_{k}b_{k}=0-0+\sum _{k=0}^{\infty }(a_{k}-a_{k+1})B_{k}}$
${\displaystyle {\begin{matrix}\sum _{k=1}^{\infty }{\frac {(-1)^{k\cdot 1}}{k}}&=&1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+{\frac {1}{5}}+...-...\\&=&1-{\frac {1}{2}}-{\frac {1}{4}}+{\frac {1}{3}}-{\frac {1}{6}}-{\frac {1}{8}}+{\frac {1}{5}}-{\frac {1}{10}}-{\frac {1}{12}}-{\frac {1}{7}}-{\frac {1}{14}}-{\frac {1}{16}}+...+{\frac {1}{2n-1}}-{\frac {1}{4n-2}}-{\frac {1}{4n}}+...\end{matrix}}}$
${\displaystyle {\frac {1}{2n-1}}-{\frac {1}{4n-2}}-{\frac {1}{4n}}={\frac {1}{4n-2}}-{\frac {1}{4n}}={\frac {1}{2}}\left({\frac {1}{2n-1}}-{\frac {1}{2n}}\right)}$

${\displaystyle S_{3n}={\frac {1}{2}}\sum _{k=1}^{n}({\frac {1}{2k-1}}-{\frac {1}{2k}}}$
${\displaystyle ={\frac {1}{2}}(1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}...+{\frac {1}{2n-1}}-{\frac {1}{2n}})}$
${\displaystyle ={\frac {1}{2}}\sum _{j=1}^{n}(-1)^{-1}{\frac {1}{j}}n\rightarrow \lim _{n\rightarrow \infty }S_{3n}}$
${\displaystyle ={\frac {1}{2}}\sum _{k=1}^{\infty }{\frac {(-1)^{k}}{k}}}$
${\displaystyle ={\frac {1}{2}}s}$
${\displaystyle S_{3n-1}=S_{3n}+{\frac {1}{2n+1}}\rightarrow {\frac {1}{2}}s}$
${\displaystyle S_{3n-2}=S_{3n+1}+{\frac {1}{4n+2}}\rightarrow {\frac {1}{2}}s}$
Siehe auch
Kriterium von Abel
Kriterium von Dirichlet

Dieser Kurs gehört zum Fachbereich Mathematik.