Use the Weierstrass M-test to show $\forall\epsilon>0,\sum_{n=1}^{\infty} a_nn^{-z}$ converges uniformly if $Re(z)>=1+\epsilon$, where $a_{n}$ is bounded.
This is what I've done: $$\sum_{n=1}^{\infty} a_nn^{-z}=\sum_{n=1}^{\infty} a_ne^{-zLog(n)}=\sum_{n=1}^{\infty} a_ne^{-(Re(z)+iIm(z))Log(n)}=\sum_{n=1}^{\infty} a_ne^{-(Re(z)+iIm(z))Log(n)}$$
I know this isn't much, but I have no idea how to handle this problem. We know $|a_{n}|<=C$ for some constant C since it is bounded. My only other idea is to say z <= R for some half disk $D=\{z:|z-c|<R\}$ for some $c > 1$. But then the issue comes with how do I get absolute values in the exponent $z$ of $\frac{1}{n^z}$?
Let $z=a+ib$ with by hypothesis $a\ge1+\epsilon$ then we have $$\left|\frac {a_n}{n^z}\right|=\left|\frac {a_n}{n^{a}}\right|\times \left|\frac {1}{n^{ib}}\right|\le\frac{M}{n^a}$$ where $M$ is an upper bound of $(a_n)$ and since the series $\sum_n \frac1{n^a}$ is a Riemann converget series then by the Weierstrass M-test the given series is convergent on the domain $$\{z\in\Bbb C\quad |\quad \operatorname{Re} z>1\}$$