The following theorem is relevant: let $\Omega$ a domain in $\mathbb{C}$ and $\{f_n\}$ a sequence of holomorphic functions in $\Omega$. If $\{f_n\}$ converges uniformly to a function $f$ in every compact subset of $\Omega$, then $f$ is holomorphic in $\Omega$ (taken from Stein and Shakarchi).
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You want to show that given $a > 0$ and $\Re(z) \ge 1 + \epsilon$ we have that $|\frac{1}{(n+a)^z}| \le \frac{C}{n^{1 + \epsilon}}$. To do this, first note that if $c$ is a real number larger than $1$, using our branch of the logarithm we have that $|c^z| = |e^{z\log(c)}| = |e^{\Re(z)\log(c)}| = c^{\Re(z)} \ge c^{1 + \epsilon}$. Thus by the sub-triangle inequality, we have that $|\frac{1}{(n + a)^z}| \le \frac{1}{(n - a)^{1 + \epsilon}}$, for large $n$.
Next, observe that since $\frac{n}{n-a} \to 1$ as $n \to \infty$, for large $n$ we have the estimate $n \le \frac{3}{2}(n-a)$. So we may take $C > 0$ to have $n \le C(n-a)$, for large $n$.
Thus: for sufficiently large $n$, we obtain the estimate $|\frac{1}{(n-a)^z}| \le C\frac{1}{n^{1 + \epsilon}}$.
From here, we deduce uniform convergence.