$\sum c_nn^{-s_0}$ converges implies $\sum c_nn^{-s}$ cannot have a pole in the line $\Re{s}=\sigma_0$ If the Dirichlet series $\sum c_nn^{-s}$ converges at $s_0=\sigma_0+it_0$, prove that the function defined by the series for $\Re{s}>\sigma_0$ cannot have a pole on the line $\Re{s}=\sigma_0$.
I hope the question meant that for any $s , \Re{s}=\sigma_0$ there is a sequence $s_k\to s$ and $M>0$ such that $|\sum c_nn^{-s_k}|<M\,\,,\forall k$
For simplicity, lets assume $s_0=0$. Then $$\sum_{n=1}^{N}c_nn^{-s}=\frac{S_N}{N^{s}}+\sum_{n=1}^{N-1}S_n\left(n^{-s}-(n+1)^{-s}\right) $$ where $S_n$ are the partial sums of the convergent series $\sum c_n$. Since $$n^{-s}-(n+1)^{-s}=sn^{-s-1}+sO(n^{-2})$$ it is enough to show the uniform boundedness along a sequence $s_k\to s$ for the sums $$\sum\frac{S_n}{n^{s_k+1}}$$ at which i am stuck. Hints or any other approach would be appreciated.

One cannot show that for every $s$ with $\operatorname{Re} s = \sigma_0$ there is a sequence $s_k \to s$ on which the sum function is bounded. For example the series
$$P(s) = \sum_{p \text{ prime}} \frac{1}{p^s}$$
converges for all $s$ with $\operatorname{Re} s \geqslant 1$, except for $s = 1$ (see e.g. here). Since
$$P(s) = \log \zeta(s) + H(s)$$
with $H$ analytic on the half-plane $\operatorname{Re} s > \frac{1}{2}$, we have
$$\lvert P(s)\rvert \sim \log \frac{1}{\lvert s-1\rvert}$$
near $1$, so $\lvert P(s_k)\rvert \to +\infty$ for every sequence $(s_k)$in $\{ s\in \mathbb{C} : \operatorname{Re} s \geqslant 1,\, s \neq 1\}$ with $s_k \to 1$.
What I know how to show is that for $s$ with $\operatorname{Re} s = s_0$ one has
$$\lim_{\varepsilon \downarrow 0}\: \varepsilon \sum_{n = 1}^{\infty} \frac{c_n}{n^{s+\varepsilon}} = 0.\tag{1}$$
This shows that the sum function doesn't have a pole at $s$, for if a meromorphic function $f$ has a pole at $s$, then one has $\lim\limits_{\varepsilon \to 0}\: \varepsilon f(s+\varepsilon) = \operatorname{Res}(f;s) \neq 0$ if the pole is simple, and $\lim\limits_{\varepsilon \to 0}\: \lvert\varepsilon f(s + \varepsilon)\rvert = +\infty$ if the order of the pole is greater than one.
To show $(1)$, we assume that $s_0 = 0$ to simplify notation (replace $c_n$ with $c_n n^{-s_0}$). Further, by modifying $c_1$, we can assume that $\sum c_n = 0$. Then, for $\operatorname{Re} s > 0$, a summation by parts gives
$$\sum_{n = 1}^{\infty} \frac{c_n}{n^s} = \sum_{n = 1}^{\infty} S_n\biggl(\frac{1}{n^s} - \frac{1}{(n+1)^s}\biggr),$$
whence
$$\sigma \Biggl\lvert\sum_{n = 1}^{\infty} \frac{c_n}{n^s}\Biggr\rvert \leqslant \sum_{n = 1}^{\infty} \lvert S_n\rvert \sigma \biggl\lvert \frac{1}{n^s} - \frac{1}{(n+1)^s}\biggr\rvert.\tag{2}$$
Now we have
$$\sigma\biggl\lvert \frac{1}{n^s} - \frac{1}{(n+1)^s}\biggr\rvert = \sigma\Biggl\lvert s\int_n^{n+1} \frac{dt}{t^{s+1}}\Biggr\rvert \leqslant \sigma \lvert s\rvert \int_n^{n+1} \frac{dt}{t^{\sigma+1}} = \lvert s\rvert\biggl(\frac{1}{n^{\sigma}} - \frac{1}{(n+1)^{\sigma}}\biggr).\tag{3}$$
By the assumption that $\sum c_n = 0$, for every $\eta > 0$ there is an $N$ such that $\lvert S_n\rvert \leqslant \eta$ for $n \geqslant N$. And there is a $C$ such that $\lvert S_n\rvert \leqslant C$ for all $n$. Inserting $(3)$ into $(2)$, and splitting the sum at $N$, we find that
\begin{align}
\sigma\Biggl\lvert \sum_{n = 1}^{\infty} \frac{c_n}{n^s}\Biggr\rvert &\leqslant \lvert s\rvert C \sum_{n = 1}^{N-1} \biggl(\frac{1}{n^{\sigma}} - \frac{1}{(n+1)^{\sigma}}\biggr) + \lvert s\rvert \eta \sum_{n = N}^{\infty} \biggl(\frac{1}{n^{\sigma}} - \frac{1}{(n+1)^{\sigma}}\biggr) \\
&= \lvert s\rvert C\biggl(1 - \frac{1}{N^{\sigma}}\biggr) + \frac{\lvert s\rvert\eta}{N^{\sigma}} \\
&\leqslant \lvert s\rvert C\biggl(1 - \frac{1}{N^{\sigma}}\biggr) + \lvert s\rvert\eta.
\end{align}
Now fix $t\in \mathbb{R}$ and let $s = \sigma + it$ with $0 < \sigma < 1$. Given $\varepsilon > 0$, choose $\eta = \frac{\varepsilon}{2(\lvert t\rvert + 1)}$, and a corresponding $N$. Then $\lvert s\rvert\eta < (\lvert t\rvert + 1)\eta = \frac{\varepsilon}{2}$. Finally, there is a $\delta \in (0,1)$ such that
$$1 - \frac{1}{N^{\sigma}} < \frac{\varepsilon}{2(\lvert t\rvert + 1)C}$$
for $\sigma \leqslant \delta$, which shows
$$\sigma \Biggl\lvert \sum_{n = 1}^{\infty} \frac{c_n}{n^{\sigma + it}}\Biggr\rvert < \varepsilon$$
for $0 < \sigma \leqslant \delta$, and the proof of $(1)$ is complete.
