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