Absolute convergence of ordinary Dirichlet series

Question

I am currently reading Serre's 'A course in Arithmetic' and I have a question about proposition 8 of section 2.4 (but I think the question can be answered without knowing the book.) The proposition says that for an ordinary Dirichlet series, $s \in \mathbb{Z}$, $f(s)=\sum\limits_{n=1}^{\infty} \frac{a_n}{n^s}$, \begin{align*} \textrm{If the } a_n \textrm{ are bounded, there is an absolute convergence for } R(s)>1. \end{align*} Serre says that this comes from the wellknown convergence of $\sum\limits_{n=1}^{\infty}\frac{1}{n^x}$ for $x >1$. Now I can see that if $a_n$ is bounded by $A$ and $s=x+iy$ we get: \begin{align*} \sum\limits_{n=1}^{\infty} \frac{a_n}{n^s}\leq A\sum\limits_{n=1}^{\infty} \frac{1}{n^s} = A\sum\limits_{n=1}^{\infty} \frac{1}{n^x}\frac{1}{n^{iy}} \end{align*} But I do not see how we get to convergence here, somehow I need to be able to say something about the imaginary part.

Answer 1

The easiest method takes a slightly different tact after your first inequality: $$A\sum \frac{1}{n^s} \leq A\sum\frac{1}{|n^s|}\leq A\sum\frac{1}{n^{|R(s)|}}$$ The last inequality holds because $|z|\geq |R(z)|$ for any complex number $z$.