Mathematik-Online problems: Problem 101: Limits of Series

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	Mathematik-Online problems:
	Problem 101: Limits of Series

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

Show that the following series have the given limits:

a) ${\displaystyle \sum_{n=1}^{\infty}\: \frac{(-3)^n}{4^n} \;=\; -\frac{3}{7} }$ b) ${\displaystyle \sum_{n=0}^{\infty}\: \frac{3^n + (-2)^n}{6^n} \;=\; \frac{11}{4} }$

c) ${\displaystyle \sum_{n=1}^{\infty}\: \frac{1}{4n^2 - 1} \;=\; \frac{1}{2} }$ d) ${\displaystyle \sum_{n=1}^{\infty}\: \frac{1}{n\,(n+1)\,(n+2)} \;=\; \frac{1}{4} }$

e) ${\displaystyle \sum_{n=0}^{\infty}\: \left(\frac{1+{\rm {i}}}{2}\right)^n \;=\; 1+{\rm {i}} }$ f) ${\displaystyle \sum_{n=1}^{\infty}\: \frac{\cos n\pi}{n} \;=\; -{\rm {ln}}\,2 }$

(Authors: Werner/Höfert)

see also:

Keyword: Limit

[Solutions]

automatisch erstellt am 14. 10. 2004

a)	${\displaystyle \sum_{n=1}^{\infty}\: \frac{(-3)^n}{4^n} \;=\; -\frac{3}{7} }$	b)	${\displaystyle \sum_{n=0}^{\infty}\: \frac{3^n + (-2)^n}{6^n} \;=\; \frac{11}{4} }$
c)	${\displaystyle \sum_{n=1}^{\infty}\: \frac{1}{4n^2 - 1} \;=\; \frac{1}{2} }$	d)	${\displaystyle \sum_{n=1}^{\infty}\: \frac{1}{n\,(n+1)\,(n+2)} \;=\; \frac{1}{4} }$
e)	${\displaystyle \sum_{n=0}^{\infty}\: \left(\frac{1+{\rm {i}}}{2}\right)^n \;=\; 1+{\rm {i}} }$	f)	${\displaystyle \sum_{n=1}^{\infty}\: \frac{\cos n\pi}{n} \;=\; -{\rm {ln}}\,2 }$