Proof: If convergence of Dirichlet series in one point, then uniform convergent in a sector I'm currently reading the proof the theorem: if a Dirichlet serie converges at some point, $s_0$, then the serie is uniformly convergent in a sector around that point. (Montgomery and Vaughan: Multiplicative Number Theory I, Thr 1.1).
But there are small things I don't understand.
So, first we define $R(u)=\sum_{n>u}a_n n^{-s_0}$ to be the remainter term of a Dirichlet serie. We note that $a_n=(R(n-1)-R(n))n^{s_0}$. Then the proof say: "so by partial summation:
\begin{align}
\sum^{N}_{n=M+1}a_nn^{-s}&=\sum^{N}_{n=M+1}(R(n-1)-R(n))n^{s_0-s}\\
&=R(M)M^{s_0s}-R(N)N^{s_0-s}-\sum^{N}_{n=M+1}R(n-1)((n-1)^{s_0-s}-n^{s_0-s})"
\end{align}
Maybe someone can explain why the last equal-sign holds.
The next question is how I can justify this bound:
$$\int^{\infty}_{M} u^{\sigma_0 - \sigma -1} \,du \leq \frac{1}{\sigma - \sigma_0}$$
when $|R(u)|\leq\epsilon$ for all $u\geq M$ and if $\sigma>\sigma_0$?
 A: It is possible to skip that summation by parts. Similar to what the book did, let's define
$$
R(u)=\sum_{n>u}{a_n\over n^{s_0}}
$$
where $s_0$ is chosen such that the series
$$
\alpha(s)=\sum_{n=1}^\infty{a_n\over n^s}
$$
converges. Now, let's pick two positive integers $M<N$, then by Riemann-Stieltjes integration (a detailed account is available in section 7.8 of Tom M. Apostol's Mathematical Analysis), we have
$$
\begin{aligned}
\sum_{n=M+1}^N{a_n\over n^s}
&=\sum_{M<n\le N}{a_n\over n^{s_0}}{1\over n^{s-s_0}}=\int_M^Nx^{s_0-s}\mathrm d[\alpha(s_0)-R(x)] \\
&=[\alpha(s_0)-R(x)]x^{s_0-s}|_M^N-\int_M^N[\alpha(s_0)-R(x)]\mathrm d(x^{s_0-s}) \\
&=R(M)M^{s_0-s}-R(N)N^{s_0-s}+\color{red}{\alpha(s_0)(N^{s_0-s}-M^{s_0-s})} \\
&-\color{blue}{\int_M^N\alpha(s_0)\mathrm d(x^{s_0-s})}+\int_M^NR(x)\mathrm d(x^{s_0-s})
\end{aligned}
$$
It can be verified that the red and the blue components cancelled each other out, leaving us
$$
\sum_{n=M+1}^N{a_n\over n^s}=R(M)M^{s_0-s}-R(N)N^{s_0-s}+\int_M^NR(x)\mathrm d(x^{s_0-s})
$$
For the second question, we have
$$
\int_M^\infty x^{\sigma_0-\sigma-1}\mathrm dx=\left.{x^{\sigma_0-\sigma}\over\sigma_0-\sigma}\right|_M^\infty={M^{\sigma_0-\sigma}\over\sigma-\sigma_0}
$$
When $\sigma>\sigma_0$ there is $\sigma_0-\sigma<0$, which implies $M^{\sigma_0-\sigma}\le1$, leaving us the following bound:
$$
\int_M^\infty x^{\sigma_0-\sigma-1}\mathrm dx\le{1\over\sigma-\sigma_0}
$$
