I am a graduate student of Mathematics. I have started reading number theory. I encountered a problem of analytic number theory.
Show that $\sum\limits_{n\in \mathbb N} \frac{2^{\omega(n)}}{n^s}=\frac{\zeta^2(s)}{\zeta(2s)}$, where $\omega(n)$ is the number of distinct prime divisors of $n$.
I have started by showing that $\omega(mn)=\omega(m)+\omega(n)$ for $(m,n)=1$. Which implies that $f(n)=2^{\omega(n)}$ is multiplicative.
I don't know what to do next. Anyone has a clue?
Best Answer
The easiest way is probably in the comment from Mastrem. But it's not hard to do it using Euler's product:
We have $$\sum_{n\ge 1} \frac{2^{\omega(n)}}{n^s} = \prod_p \sum_{k\ge 0}\frac{2^{\omega(p^k)}}{p^{ks}}= \prod_p \left(1+\sum_{k\ge 1}\frac{2}{p^{ks}}\right)$$
$$= \prod_p \left(1+\frac{2}{p^s-1}\right)= \prod_p \frac{p^s+1}{p^s-1}= \prod_p \frac{p^{2s}-1}{(p^s-1)^2}$$
$$= \prod_p \left(\frac{p^s}{p^s-1}\right)^2 \frac{p^{2s}-1}{p^{2s}} = \frac{\zeta(s)^2}{\zeta(2s)}$$