# If $(a_n)_n$ is bounded, then $\sum\limits_n a_{n}n^{-z}$ converges uniformly for $\Re z \geq 1+\epsilon$

More specifically, let $(a_n)_n$ be a bounded sequence of complex numbers. Show that for each $\epsilon>0$, the series $\sum\limits_{n=0}^{\infty} a_{n}n^{-z}$ converges uniformly for $\Re z \geq 1+\epsilon$, choosing the principal branch of $n^{-z}$.

It seems like we can bound the series by $B \sum\limits_{n=0}^{\infty} n^{-z}$, but how do we deal with the complex square root?

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Hint: what is $|n^{-z}|$? –  Did Nov 8 '12 at 7:32
Rewrote title and parts of the post. –  Did Nov 8 '12 at 7:41

Hint: use what @did says + Weierstrass M-test + $\sum_{k=1}^{\infty}\frac{1}{k^{\alpha}}$ is finite when $\alpha>1$.