Show that $f(s)=\sum \frac{1}{n^s}$ is continuous for $Re(s)>1$

Question

Show that $f(s)=\sum \frac{1}{n^s}$ is continuous for $Re(s)>1$. In my attempt i try to use Weierstrass Test, $\frac{1}{n^s}=\frac{1}{n^{Re(s)+iIm(s)}}=\frac{1}{e^{\log n^{\Re(s)+iIm(s)}}}=\frac{1}{e^{Re(s)\log n}e^{iIm(s)\log n}}$ then $|\frac{1}{n^s}|=\frac{1}{n^{Re(s)}}$ and $|\sum \frac{1}{n^s}|\leq \sum \frac{1}{n^{Re(s)}}<\infty$ and by Weierstrass Test the series converge uniformly and just i need (Im not sure) say that every function $\frac{1}{n^s}$ is Continuous? For conclude that the series is continuous? In my way im wront ? I want to finish this exercise please!! Thank you

Answer 1

Take $s\in\Bbb C$ such that $\operatorname{Re}(s)>1$. Now, take $r\in\left(1,\operatorname{Re}(s)\right)$. For each $z\in\Bbb C$ such that $\operatorname{Re}(z)>r$, you have$$\sum_{n=1}^\infty\left|\frac1{n^z}\right|=\sum_{n=1}^\infty\frac1{n^{\operatorname{Re}(z)}}\leqslant\sum_{n=1}^\infty\frac1{n^{\operatorname{Re}(s)}}.$$So, by the Weierstrass $M$-test, and since each function $z\mapsto\frac1{n^z}$ is continuous, the function $z\mapsto\zeta(z)=\sum_{n=1}^\infty\frac1{n^z}$ is continuous on $\{z\in\Bbb C\mid\operatorname{Re}(z)>r\}$. Since this set is open and $s$ belongs to it, the function $\zeta$ is continuous at $s$.