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


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.

  • $\begingroup$ 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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