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

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

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

2. Originally Posted by silversand
$\displaystyle \sum_{n=1}^{\infty}{\frac{1}{n^{x+1}}}$
for it not to be continuous then

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

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

which is not on $\displaystyle (0,1]$

therefore must be continuous

3. The function can be written as...

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

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$