Question about the covergence of a Dirichlet series

Question

Suppose that F($s$) = $\sum \frac{a(n)}{n^s}$ is a Dirichlet series, where the sum is taken for all intergers $n\geq 1$. It's also known that the series converges for all complex numbers $s$ with $Re(s)>1$. What I'm trying to show is that the limit $\frac{A(x)}{x^s}$ as $x\rightarrow\infty$ is 0, where $A(x)=\sum a(n)$ with the sum being taken over all $n\leq x$. I've tried using various properties of convergence of Dirichlet series and even tried using the Abel summation, but to no avail. Can someone please give a solution?

Answer 1

It doesn't sound like you've tried very hard if you used Abel summation but couldn't get there! We have that \[A(x) = x^s \sum_{n \leq x} \frac{a(n)}{n^s} - s \int_1^x t^{s - 1} \sum_{n \leq t} \frac{a(n)}{n^s} dt.\] Now as $\sum_{n = 1}^{\infty} \frac{a(n)}{n^s}$ converges, this is equal to \begin{align} A(x) & = x^s \left(\sum_{n = 1}^{\infty} \frac{a(n)}{n^s} + o(1)\right) - s \int_1^x t^{s - 1} \left(\sum_{n = 1}^{\infty} \frac{a(n)}{n^s} + o(1)\right) dt\\ & = \left(x^s - s\int_1^x t^{s - 1} dt\right) \sum_{n = 1}^{\infty} \frac{a(n)}{n^s} + o\left(x^{\Re(s)} + |s| \int_1^x t^{\Re(s) - 1} dt\right)\\ & = \sum_{n = 1}^{\infty} \frac{a(n)}{n^s} + o\left(\frac{|s|}{\Re(s)} x^{\Re(s)}\right), \end{align} and so \[\frac{A(x)}{x^s} = o(1).\]