Show that $\sum\limits_{n\in \mathbb N} \frac{2^{\omega(n)}}{n^s}=\frac{\zeta^2(s)}{\zeta(2s)}$.

Question

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?

Answer 1

By definition, we know

$$ 2^{\omega(n)}=\prod_{p|n}(1+1)=\sum_{d|n}\mu^2(d) $$

Moreover, it can be verified that

$$ \mu^2(d)=\sum_{d^2|n}\mu(d) $$

This suggests that

\begin{aligned} \sum_{n\ge1}{\mu^2(n)\over n^s} &=\sum_{n\ge1}{1\over n^s}\sum_{d^2|n}\mu(d)=\sum_{d\ge1}\mu(d)\sum_{\substack{n\ge1\\d^2|n}}{1\over n^s} \\ &=\sum_{d\ge1}\mu(d)\sum_{k\ge1}{1\over(d^2k)^s}=\zeta(s)\sum_{d\ge1}{\mu(d)\over d^{2s}} \end{aligned}

By the properties of Möbius inversion, it is evident that $\sum_{n\ge1}\mu(n)n^{-s}=1/\zeta(s)$, so using the properties of Dirichlet convolution we have

$$ \sum_{m\ge1}{2^{\omega(m)}\over m^s}=\zeta(s)\sum_{n\ge1}{\mu^2(n)\over n^s}={\zeta^2(s)\over\zeta(2s)} $$

Answer 2

The easiest way is probably in the comment from Mastrem. But it's not hard to do it using Euler's product:

We have $$\sum_{n\ge 1} \frac{2^{\omega(n)}}{n^s} = \prod_p \sum_{k\ge 0}\frac{2^{\omega(p^k)}}{p^{ks}}= \prod_p \left(1+\sum_{k\ge 1}\frac{2}{p^{ks}}\right)$$

$$= \prod_p \left(1+\frac{2}{p^s-1}\right)= \prod_p \frac{p^s+1}{p^s-1}= \prod_p \frac{p^{2s}-1}{(p^s-1)^2}$$

$$= \prod_p \left(\frac{p^s}{p^s-1}\right)^2 \frac{p^{2s}-1}{p^{2s}} = \frac{\zeta(s)^2}{\zeta(2s)}$$