Is the function $f: (1, \infty) \to \mathbb R$ defined as $f(x)=\sum_{n=1}^\infty n^{-x}$ continuous ? I know that for each
$n \ge 1$ , the function $g:(1,\infty) \to \mathbb R ; g(x)=n^{-x}$ is uniformly continuous . So for a given $\epsilon >0 , a \in (1,\infty) , n \ge 1 , \exists \delta_{\epsilon,n}$ such that $|x-a|<\delta _{\epsilon , n} \implies |n^{-x}-n^{-a}|<\dfrac \epsilon{2^n}$ but if we then want to sum over $n$ , to make each $\epsilon /2^n$ estimate valid , we have to take
$\max\{\delta_{\epsilon , n} : n \ge 1\}$ which may not exist as we are taking maximum over a set which we don't know is finite or not . So I am stuck . Please help .
For every $b>a>1$ let the functions $f_n:[a,b]\to\mathbb R$ be defined by $f_n=n^{-x}$, then these functions are all continuous on $[a,b]$ and are bounded by $\frac1{n^a}$ on $[a,b]$, so by Weierstrass's M-test for uniform convergence of series, and by convergence of $\sum_{n=1}^\infty \frac1{n^a}$ we find out that the function $\sum_{n=1}^\infty \frac1{n^x}$ is continuous on every closed sub interval of $(1,\infty)$ and so continuous on the whole of interval!