I'm trying to figure out how to show that $$\sum_{n=1}^{\infty} \frac{1}{n^z}$$ converges locally uniformly on the set $\textrm{Re}(z) > 1$ The only ideas I've had are trying to use the Weierstrass M-Test but I can't think of any $M_n$ that would apply and trying to show that the sequence is uniformly Cauchy on said set, but made no progress with that either.
Thank you!
Take a compact set $K$ contained in $\{\operatorname{Re}(z) > 1\}$; then there exists $\epsilon > 0$ such that $$z \in K \implies \operatorname{Re}(z) > 1 + \epsilon.$$
Now we have
$$\sum_{n = 1}^{\infty} \frac 1 {|n^z|} = \sum_{n = 1}^{\infty} \frac{1}{n^{1 + \epsilon}} < \infty$$
and the Weierstrass $M$-test applies.