I am currently studying number theory and our instructor refers to Apostol's book on Analytic number theory for the chapter Dirichlet series.In that book,there is an exercise which is as follows:
Let $d$ be the divisor function $\sigma_0$.Then we have to show that the following equation holds for $Re(s)>1$
$\sum\limits_{n=1}^\infty \frac{d(n^2)}{n^s}=\frac{\zeta^3(s)}{\zeta(2s)}$
I am a beginner and I have no idea how to solve this.I want some help.Can someone write a detailed solution for me?I was trying to proceed using Euler product formula as $d(n^2)$ is multiplicative.Then I get something like:
$\sum\limits_{n=1}^\infty \frac{d(n^2)}{n^s}=\prod\limits_{p}(1+\frac{3}{p^s}+\frac{5}{p^{2s}}+\dots)$.But then I do not know what to do next.I am not also able to show that the sum converges absolutely for $Re(s)>1$.
Addendum
I have used that $\sum\limits_{n=0}^\infty (2n+1)x^n=\frac{x+1}{(1-x)^2}$.
Thus we get $\sum\limits_{n=1}^\infty \frac{d(n^2)}{n^s}=\prod\limits_{p}\frac{p^s(1+p^s)}{(1-p^s)^2}=\frac{\zeta^3(s)}{\zeta(2s)}$ by using $\zeta(s)=\prod\limits_{p}(1-p^{-s})^{-1}$ but still I don't know why the series converges absolutely for $Re(s)>1$.
Best Answer
Notice that $d(n^2)<n^2,$ and so you can write the Euler product for the Dirichlet series in $\Re s > 3.$ Your computations show that $\sum_{n=1}^\infty \frac{d(n^2)}{n^s} = \frac{\zeta^3(s)}{\zeta(2s)}$ in that region. Now the right hand side converges absolutely in $\Re s >1$ and so must the left hand side, by uniqueness of Dirichlet series.