# Dirichlet series and Riemann zeta function

Im trying to show, for $\Re(s)>1$, that $\displaystyle\sum_{n=0}^{\infty} \frac{d(n^2)}{n^s} = \frac{\zeta^3(s)}{\zeta(2s)}$, where $d(n)= |\{k \mid k|n \}|$, number of positive integers that divides $n$.

I tried to separate the RHS to $\displaystyle{\zeta^2(s)}=\sum_{n=0}^{\infty} \frac{u* u(n)}{n^s}=\sum_{n=0}^{\infty} \frac{d(n)}{n^s}$ which obtained by the dirichlet sum of $\zeta (s)$ and $\dfrac{\zeta(s)}{\zeta(2s)}$ which I tried to simplify by Euler Product formula but it didnt came up to something.

The Euler product of $\zeta(s)^3/\zeta(2s)$ is $$\frac{\zeta(s)^3}{\zeta(2s)} = \prod_p \frac{1-p^{-2s}}{(1-p^{-s})^3} = \prod_p \frac{1+p^{-s}}{(1-p^{-s})^2}.$$ But $$\frac{1+p^{-s}}{(1-p^{-s})^2} = \sum_{k=0}^n (2k+1)p^{-ks} = \sum_{k=0}^n d(p^{2k})p^{-ks}.$$ Since $d$ is multiplicative, you should be able to conclude.
• Could you please explain the transition from $$\frac{1+p^{-s}}{(1-p^{-s})^2}$$ to the infinite sum? – Snufsan Jul 30 '14 at 12:58
• Still i cant figure how we got the $2k+1$ factor. we have $$\frac{1}{(1-p^{-s})^2} = \sum_{n=0}^{\infty} \frac{d(p^k)}{p^{ks}}$$ and then what? – Snufsan Jul 30 '14 at 13:21