Uniform convergence of Dirichlet series Let $a_1,a_2,\ldots\in\mathbb{C}$ and consider the Dirichlet series $\sum_{n=1}^\infty \dfrac{a_n}{n^z}$. Suppose the series converges for some $z_0$. Then why does it converge uniformly on every closed disk in the half-space $\Re z>\Re z_0$, where $\Re$ denotes the real part of a complex number?
I think it's reasonable that when the real part of $z$ increases, the denominator of $\dfrac{a_n}{n^z}$ increases (well, in some sense), so the convergence gets better. How can we actually prove it though?
 A: For simplifying notation, let
$$S_k=\sum_{m\geq k}\frac{a_n}{n^{z_0}}$$
Now, we do the classical Abel's trick and for all $N,M\in \mathbb{N}$ with $N<M$, we obtain
\begin{align*}
\sum_{n=N}^M\frac{a_n}{n^z}&=\sum_{n=N}^M\frac{a_n}{n^{z_0}}\frac{1}{n^{z-z_0}}\\
&=\sum_{n=N}^M\frac{S_{n+1}-S_{n}}{n^{z-z_0}}\\
&=\sum_{n=N}^M\frac{S_{n+1}}{n^{z-z_0}}-\sum_{n=N}^M\frac{S_{n}}{n^{z-z_0}}\\
&=\sum_{n=N+1}^{M}S_n\left(\frac{1}{(n-1)^{z-z_0}}-\frac{1}{n^{z-z_0}}\right)+\frac{S_{M+1}}{M^{z-z_0}}-\frac{S_N}{N^{z-z_0}}
\end{align*}
after rutinary operations with each sumatory. Here, we have to observe that
\begin{align*}
\left|\frac{1}{(n-1)^{z-z_0}}-\frac{1}{n^{z-z_0}}\right|=\left|\int_{n-1}^{n}\frac{z-z_0}{x^{z-z_0+1}}\,dx\right|&\\\leq |z-z_0| \int_{n-1}^{n}\frac{dx}{x^{\text{Re}(z-z_0)+1}}&=\frac{|z-z_0|}{\text{Re}(z-z_0)}\left(\frac{1}{(n-1)^{\text{Re}(z-z_0)}}-\frac{1}{n^{\text{Re}(z-z_0)}}\right)
\end{align*}
under the assumption $\text{Re}(z-z_0)>0$. Therefore for $\text{Re}(z-z_0)>0$ and $M>N$ we obtain
$$\left|\sum_{n=N}^M\frac{a_n}{n^z}\right|\leq \frac{\max_{N\leq n\leq M+1}|S_n|}{N^{\text{Re}(z-z_0)}}\left(2+\frac{|z-z_0|}{\text{Re}(z-z_0)}\right)$$
after seeing that
\begin{align*}
\left|\sum_{n=N}^M\frac{a_n}{n^z}\right|&=\sum_{n=N+1}^{M}|S_n|\left|\frac{1}{(n-1)^{z-z_0}}-\frac{1}{n^{z-z_0}}\right|+\frac{|S_{M+1}|}{M^{\text{Re}(z-z_0)}}+\frac{|S_N|}{N^{\text{Re}(z-z_0)}}\\
&\leq \left(\max_{N<n\leq M}|S_n|\right)\sum_{n=N+1}^{M}\left|\frac{1}{(n-1)^{z-z_0}}-\frac{1}{n^{z-z_0}}\right|+\frac{|S_{M+1}|}{M^{\text{Re}(z-z_0)}}+\frac{|S_N|}{N^{\text{Re}(z-z_0)}}\text{;}
\end{align*}
\begin{align*}
\sum_{n=N+1}^{M}\left|\frac{1}{(n-1)^{z-z_0}}-\frac{1}{n^{z-z_0}}\right| &\leq \frac{|z-z_0|}{\text{Re}(z-z_0)}\sum_{n=N+1}^{M}\left(\frac{1}{(n-1)^{\text{Re}(z-z_0)}}-\frac{1}{n^{\text{Re}(z-z_0)}}\right)\\
&\leq \frac{|z-z_0|}{\text{Re}(z-z_0)}\left(\frac{1}{N^{\text{Re}(z-z_0)}}-\frac{1}{M^{\text{Re}(z-z_0)}}\right)\\
&\leq \frac{|z-z_0|}{\text{Re}(z-z_0)}\frac{1}{N^{\text{Re}(z-z_0)}}\text{,}
\end{align*}
since $\sum_{n=N+1}^{M}\left(\frac{1}{(n-1)^{\text{Re}(z-z_0)}}-\frac{1}{n^{\text{Re}(z-z_0)}}\right)$ is a telescoping series; and that
\begin{align*}
\left(\max_{N<n\leq M}|S_n|\right)\frac{|z-z_0|}{\text{Re}(z-z_0)}\frac{1}{N^{\text{Re}(z-z_0)}}&+\frac{|S_{M+1}|}{M^{\text{Re}(z-z_0)}}+\frac{|S_N|}{N^{\text{Re}(z-z_0)}}\\&\leq \frac{\max_{N\leq n\leq M+1}|S_n|}{N^{\text{Re}(z-z_0)}}\left(2+\frac{|z-z_0|}{\text{Re}(z-z_0)}\right)\text{,}
\end{align*}
due to $N<M$.
Now, we are ready for the final step, but before three remarks:


*

*Since $\text{Re}(z-z_0)>0$,
$$\frac{1}{N^{\text{Re}(z-z_0)}}<1\text{.}$$

*Since every closed ball $B$ is compact and if contained in the
considered region, the function  $$z\mapsto
    \frac{|z-z_0|}{\text{Re}(z-z_0)}$$ is well defined and continuous
over it, there is a constant $K_B$ such that for all $z\in B$,
$$\frac{|z-z_0|}{\text{Re}(z-z_0)}\leq K_B\text{.}$$ Note that here
the important part is that the considered function is bounded, for
which we can take far bigger domains -e.g., $A(z_0,\theta)= \{z\in
    \mathbb{C}\,|\,\text{Re}(z)>\text{Re}(z_0)\text{ and
    }\text{arg}(z-z_0)\leq \theta\}$ for $\theta<\frac{\pi}{2}$-.

*Since $\sum \frac{a_n}{n^{z_0}}$ converges, $S_K$ converges for zero
and thus for all $\varepsilon>0$, there is $C_\varepsilon>0$ such
that  $$|S_n|<\varepsilon$$ whenever $n\geq C_\varepsilon$.


Joining this three remarks and the previous formula, we obtain that given a close ball $B$ contained in the half plane $\text{Re}(z)>\text{Re}(z_0)$ and $\varepsilon>0$, for all $z\in B$ and $N,M\geq C_{\varepsilon/(2+K_B)}$,
$$\left|\sum_{n=N}^M\frac{a_n}{n^z}\right|<\varepsilon\text{.}$$
Thus $\sum_{n=1}^\infty\frac{a_n}{n^z}$ is uniformly Cauchy in $B$ and therefore converges uniformly in B as desired.
A: Apply what you know about complex series:
$$\text{Re}\,z>\text{Re}\,z_0\implies \sum_{n=1}^\infty\left|\frac{a_n}{n^z}\right|=\sum_{n=1}^\infty\frac{|a_n|}{n^{\text{Re}\,z}}\le\sum_{n=1}^\infty\frac{|a_n|}{n^{\text{Re}\,z_0}}=\sum_{n=1}^\infty\left|\frac{a_n}{n^{z_0}}\right|$$
and since the rightmost series converges then also ours does
