Analytic continuation of Dirichlet function Suppose $\{a_n\}$ is a sequence of complex numbers such that the sums $A_n=a_1+\cdots+a_n$ satisfy $$|A_n-nb|\leq Cn^{\sigma}$$ for all $n$, where $b\in\mathbb{C},C>0,0\leq\sigma<1$. Prove that the function $f(z)$ defined by $$f(z)=\sum_{n=1}^\infty\dfrac{a_n}{n^z}$$ for $\Re{z}>1$ has an analytic continuation to the region $\Re{z}>\sigma$ except for a simple pole of residue $b$ at $z=1$. ($\Re$ denotes the real part of a complex number).
I want to consider $f(z)-b\zeta(z)$ and use the theorem that $\zeta(z)$ can be extended to an analytic function on $\Re z>0$ except for a simple pole at $z=1$.
 A: We have $f(z) - b\zeta(z) = \sum_{n=1}^\infty \frac{c_n}{n^z}$ where $c_n=a_n-b$. The assumption gives you $|c_1+\cdots+c_n| \le C n^\sigma$. You should have a theorem available to you that tells you that under that condition, $\sum_{n=1}^\infty \frac{c_n}{n^z}$ converges for all $z$ with $\Re z>\sigma$, which would solve your problem.
To do it by hand, let $T(n) = c_1 + \cdots + c_n$. Then for real $x>\sigma$,
$$
\sum_{n=1}^\infty \frac{c_n}{n^x} = \sum_{n=1}^\infty \frac{T(n)-T(n-1)}{n^x} = \sum_{n=1}^\infty T(n) \bigg( \frac1{n^x} - \frac1{(n+1)^x} \bigg).
$$
By the mean value theorem,
$$
\frac1{n^x} - \frac1{(n+1)^x} = (n-(n-1))\frac{-x}{\eta^{x+1}} = \frac x{\eta^{x+1}}
$$
for some $\eta\in [n,n+1]$; in particular, it is at most $x/n^{x+1}$. Therefore
$$
\bigg| \sum_{n=1}^\infty \frac{c_n}{n^x} \bigg| \le \sum_{n=1}^\infty Cn^\sigma \frac x{n^{x+1}} = Cx \sum_{n=1}^\infty \frac1{n^{x+1-\sigma}},
$$
which converges when $x\ge\sigma$. This proves the desired statement when $z$ is real, but you should have access to a theorem that deduces the full statement (convergence in the right-half plane) from that.
