I am running into considerable trouble trying to prove the identity in the question. I figure the solution will come from Euler-products, so here was my attempt. I want to show that
$$
\sum_{n=1}^\infty \frac{d(n)^2}{n^s} = \frac{\zeta(s)^4}{\zeta(2s)},
$$
where
$$
d(n):=\sum_{d\mid n} 1.
$$
First for prime $p$, $d(p)=2$ and $d(p^n)=n+1$,
$$
\begin{align*}
\sum_{n=1}^\infty \frac{d(n)^2}{n^s} &= \prod_p \left(1 + \frac{d(p)^2}{p^s} + \frac{d(p^2)^2}{p^{2s}} + \frac{d(p^3)^2}{p^{3s}} + \cdots\right) \\
&= \prod_p \left(1+\frac{2^2}{p^s} + \frac{3^2}{p^{2s}}+\frac{4^2}{p^{3s}}+\cdots \right) \\
&= \prod_p \sum_{n=1}^\infty \frac{n^2}{p^{(n-1)s}} \\
&= \prod_p \frac{p^{2s}(p^s+1)}{(p^s-1)^3}, \qquad \text{according to Wolfram Alpha}
\end{align*}
$$
Here is where I get stuck. My next thought was if we look at what the Euler-product form of $\frac{\zeta(s)^4}{\zeta(2s)}$ we might be able to see the connection.
$$
\frac{\zeta(s)^4}{\zeta(2s)} = \prod_p\frac{\left(1-\frac1{p^s}\right)^4}{\left(1-\frac1{p^{2s}}\right)} = \prod_p \frac{p^{-4s}-4p^{-3s}+6p^{-2s}-4p^{-s}+1}{1-p^{-2s}}
$$
but I see no way to transform this product into the one we have derived.
From here I have no clue how to proceed. I also know that Ramanujan's identity
$$
\sum_{n=1}^\infty \frac{\sigma_a(n)\sigma_b(n)}{n^s}=\frac{\zeta(s)\zeta(s-a)\zeta(s-b)\zeta(s-a-b)}{\zeta(2s-a-b)}
$$
would solve this for me, but I'd like to do it directly for this specific case to get used to manipulating products like this. Any hints on how to continue?
Best Answer
Actually, $$\frac{\zeta(s)^4}{\zeta(2s)} = \prod_p\frac{\left(1-\frac1{p^{2s}}\right)} {\left(1-\frac1{p^s}\right)^4}.$$ Then $$\prod_p\frac{\left(1-\frac1{p^{2s}}\right)} {\left(1-\frac1{p^s}\right)^4} =\frac{p^{2s}(p^{2s}-1)}{(p^s-1)^4}=\frac{p^{2s}(p^s-1)(p^s+1)}{(p^s-1)^4} =\frac{p^{2s}(p^s+1)}{(p^s-1)^3} $$ etc.