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I am stuck on how to do this question:

Let d(n) denote the number of divisors of n. Show that the dirichlet series generating function of the sequence {(d(n))^2} equals C^4 (s)/ C(2s).

C(s) represents the riemann zeta function, I apologize, I am not very accustomed with LaTex. Any help is highly appreciated, I am studying for an exam. Everywhere I have looked on the internet says it is obvious, but none of them seem to want to explain why or how to do it, so please help. thank you

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There are many different ways to approach this, depending on what you are permitted to use. One simple way is to use Euler products.

The Euler product for $$Q(s) = \sum_{n\ge 1} \frac{d(n)^2}{n^s}$$ is given by $$ Q(s) = \prod_p \left( 1 + \frac{2^2}{p^s} + \frac{3^2}{p^{2s}} + \frac{4^2}{p^{3s}} + \cdots \right).$$ This should follow by inspection considering that $d(n)$ is multiplicative and $d(p^v) = v+1$, with $p$ prime.

Now note that $$\sum_{k\ge 0} (k+1)^2 z^k = \sum_{k\ge 0} (k+2)(k+1) z^k - \sum_{k\ge 0} (k+1) z^k \\= \left(\frac{1}{1-z}\right)'' - \left(\frac{1}{1-z}\right)' = \frac{1+z}{(1-z)^3}.$$

It follows that the Euler product for $Q(s)$ is equal to $$ Q(s) = \prod_p \frac{1+1/p^s}{(1-1/p^s)^3}.$$

On the other hand, we have $$ \zeta(s) = \prod_p \frac{1}{1-1/p^s}$$ so that $$ \frac{\zeta^4(s)}{\zeta(2s)} = \prod_p \frac{\left(\frac{1}{1-1/p^s}\right)^4}{\frac{1}{1-1/p^{2s}}} = \prod_p \frac{1-1/p^{2s}}{\left(1-1/p^s\right)^4} = \prod_p \frac{1+1/p^s}{\left(1-1/p^s\right)^3}.$$ The two Euler products are the same, QED.

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  • $\begingroup$ Thank you very much, I see now, you put this answer in a very logical way that was easy to follow. I am new to this site and if possible to give you points/badge i would like to do so, I am just unsure how? Thanks again $\endgroup$ – moony May 9 '13 at 14:49

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