A certain identity of a Dirichlet series

Question

I have encountered this problem: I need to prove that $\sum_{n=1}^{\infty} \frac{d(n^2)}{n^s} = \frac{\zeta^{3}(s)}{\zeta(2s)}$. Now, I already know that $\frac{\zeta(s)}{\zeta(2s)} = \sum_{n is square free} \frac{1}{n^s} = \sum_{n=1}^{\infty} \frac{\mu^{2}(n)}{n^s} $, and $\sum_{n=1}^{\infty} \frac{d(n)}{n^s} = \zeta^{2}(s)$ , So this problem can be solved if I can prove this convulation identity: $(\mu^{2} * d)(n) = d(n^{2}) $. I tried some things but it didn't work out, someone can help\have mmaybe another way?

Answer 1

By definition, $d(n)=\sum_{d|n}1$ is multiplicative, and $\mu^2(n)$ is also some multiplicative function. Hence, we may consider the prime powers first:

Let $n=p^k$ where $k\ge1$, so $$ \begin{aligned} \sum_{m|n}\mu^2(m)d\left(\frac nm\right) &=\sum_{r=0}^k\mu^2(p^r)d(p^{k-r}) \\ &=\mu^2(1)d(p^k)+\mu^2(p)d(p^{k-1}) \\ &=d(p^k)+d(p^{k-1}) \\ &=2k+1=d(p^{2k}) \end{aligned} $$

As a result, we have $$ \sum_{m|n}\mu^2(m)d\left(\frac nm\right)=d(n^2) $$ which completes the proof.