# Verifying identities for Riemann zeta function

I ran across these two problems, while reading a text on number theory. The problem states: "Verify the following identities". What does "verify" mean in this context, and what strategies can I employ in trying to solve these questions?

1.) $\zeta(s)^{-1}= \sum^{\infty}_{n=1}\mu(n)/n^{s}$

2.) $\zeta(s)^{2}= \sum^{\infty}_{n=1}\nu(n)/n^{s}$

where $\zeta(s)=\sum^{\infty}_{n=1}1/n^{s}$ is the Riemann $\zeta$ function, $\mu(n)$ is the Möbius $\mu$ function and $\nu(n)$ counts the number of positive divisors of n.

-
Your first question was answered here. As anon explains in his answer, Dirichlet convolution is one of the simpler ways to verify your given identities. – J. M. Jan 26 '12 at 7:41

"Verify" likely means perform some kind of symbolic computation that checks that the identities are in fact true. As for strategies, there is the route of direct multiplication of Dirichlet series:

$$\left(\sum_{n=1}^\infty a_nn^{-s}\right)\left(\sum_{m=1}^\infty b_mm^{-s}\right)=\sum_{n,m=1}^\infty a_nb_m(nm)^{-s}\;\stackrel{k=nm}{=}\; \sum_{k=1}^\infty \left(\sum_{nm=k}a_nb_m\right)k^{-s}.$$

Note that the coefficients above may be rewritten as $\sum_{d|k}a_db_{k/d}$. So if you have access to the fact

$$\sum_{d|k}\mu(d)\cdot1=\begin{cases}1&k=1\\0&k>1,\end{cases}$$

then you're golden. You'll also need the fact that $v(n)=\sum_{d|n}1$, naturally. Otherwise, for the first problem anyway, you can use the Euler product, viz.

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

$$=\sum_{n\text{ squarefree}}(-1)^{\omega(n)}n^{-s}=\sum_{n=1}^\infty \mu(n)n^{-s}.$$

(This is sort of hand-wavy though.)

-