Elegent argument for the convergence of $\sum_{n = 1}^{\infty} \frac{\cos(n)}{n^z}$?

Question

I want to prove that $$ f(z) = \sum_{n = 1}^{\infty} \frac{\cos(n)}{n^z} \quad (\mathrm{Re}(z) > 1)$$ defines a holomorphic function. Obviously this would be the case if the series $$ |f(z)| \leq \sum_{n = 1}^{\infty} \frac{|\cos(n)|}{n} $$ converges, but I'm not sure that is the case.

Answer 1

By comparison:

$$\biggl\lvert\frac{\cos n}{n^z}\biggr\rvert=O\biggl(\frac1{n^{\operatorname{Re}z}}\biggr),$$ which converges if $\;\operatorname{Re}z>1$.

Answer 2

No, the series $\sum\frac{|\cos(n)|}{n}$ does not converge. (Loosely speaking, roughly half the terms are as large as $c/n$.)

Hint: Saying $f(z)$ is holomorphic in the region $\Re z>1$ is the same as saying that it's holomorphic in the region $\Re z>1+\epsilon$ for every $\epsilon>0$.