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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?

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    $\begingroup$ Are you familiar with Dirichlet convolution -- this is where I go when my denominator is a product. I note that the divisor function, $d$, has $d = 1 \ast 1$. $\endgroup$ Dec 7 '19 at 16:53
  • $\begingroup$ @EricTowers I am familiar with Dirichlet convolution, but I've only seen it in the context of multiplying two Dirichlet series. Do you mean that there is a way to use it here or that I should restart with the Dirichlet series representations of $\zeta(s)^4$ and $\frac{1}{\zeta(2s)}$ and multiply them that way? $\endgroup$
    – Logan Toll
    Dec 7 '19 at 16:57
  • $\begingroup$ Yes, prime-wise is smart. And, also, not positive powers of $p$ or $p^s$, but negative... and then things turn out as sanely as we can hope. Of course, there's Estermann's example of $d(n)^3$, which (provably) cannot be re-arranged to anything elementary. $\endgroup$ Aug 30 at 21:25
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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.

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  • $\begingroup$ Wow.... I spent the better part of an hour staring at this. Thank you... :/ $\endgroup$
    – Logan Toll
    Dec 7 '19 at 16:56

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