show this series defines a continuous function on (0,1]

**silversand** · May 3rd 2009, 07:03 PM

$\sum_{n=1}^{\infty}{\frac{1}{n^{x+1}}}$

**pickslides** · May 3rd 2009, 10:07 PM

Originally Posted by silversand

$\sum_{n=1}^{\infty}{\frac{1}{n^{x+1}}}$

for it not to be continuous then

$n^{x+1} = 0$ on the given interval

$x+1 = \log_n(0)$ is undefined!

which is not on $(0,1]$

therefore must be continuous

**chisigma** · May 3rd 2009, 10:58 PM

The function can be written as...

$f(x)= \sum_{n=1}^{\infty}\frac{1}{n^{1+x}}= \sum_{n=1}^{\infty} e^{-(1+x)\cdot \ln n}$ (1)

... so that for any $x_{0}>0$ it exists an $\epsilon$ with $0<|\epsilon|<x_{0}$ and for which ...

$\lim_{\epsilon \rightarrow 0} f(x_{0}+\epsilon)= \lim_{\epsilon \rightarrow 0} \sum_{n=1}^{\infty} e^{-(1+x_{0})\cdot \ln n}\cdot e^{-\epsilon \cdot \ln n}=$

$= \lim_{\epsilon \rightarrow 0} \sum_{n=1}^{\infty} \frac{1}{n^{1+x_{0}}}\cdot \frac{1}{n^{\epsilon}}= \lim_{\epsilon \rightarrow 0} \sum_{n=1}^{\infty} \frac{1}{n^{1+x_{0}}} = f(x_{0})$ (2)

Therefore $f(x)$ defined in (1) is a continous function...

Kind regards

$\chi$ $\sigma$

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