Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 6 down vote accepted

"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


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.


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

(This is sort of hand-wavy though.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.