I'm trying to prove the following:
$$ \frac{\zeta(s)^2}{\zeta(2s)} = \sum_{n = 1}^{\infty} 2^{\omega(n)} n^{-s} $$
So far I have:
$$ \zeta(s)^2 = \sum_{m = 1}^{\infty} \tau(m) m^{-s} $$
and of course
$$ \zeta(s)^{-1}= \sum_{n = 1}^{\infty} \mu(m) n^{-2s} $$
where $\tau(n)$ is the number of divisors of $n$ and $\mu$ is the Mobius function.
Normally you could rewrite $\frac{\zeta(s)^2}{\zeta(2s)}$ using convolutions and $L$-functions (which is how I got $\zeta(s)^2$ and $\zeta(s)^{-1}$, but $2s$ instead of $s$ complicates things. Simply working out the product yields $$ \frac{\zeta(s)^2}{\zeta(2s)} = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \mu(m) \tau(n) (m^2 n)^{-s} $$ This may be rewritten as
$$ \sum_{k = 1}^{\infty} k^{-s} \cdot \sum_{m^2 n = k} \mu(m) \tau(n) $$
Hence we have to prove that the last sum equals $\omega(k)$.
The problem is my lack of understanding of $\omega(n)$ (I understand the definition but I don't know of any identities involving it). Where does the $2^{\omega(n)}$ come from? Is this even the right approach?
Any help is greatly appreciated.
P.S.: Don't spoil it please! I'd prefer a few simple hints.
I think you have chosen a complicated approach. There is a much simpler way. You know the Euler product for $\zeta(s)$, and thus you know the Euler product for the left hand side. The function $2^{\omega(n)}$ is multiplicative and easy to understand on prime powers, so you should write down the Euler product for the right hand side. Then you can verify that they are equal.
This is what I think of as the canonical simple approach to problems of this sort. The crazy convolution identity that results from this is then one of the fruits of the labour.