I came across this question.
Let $d(n)$ denote the number of divisors of $n$. Let $\nu(z) = \sum\limits_{n=1}^{\infty} d(n) z^{n}$ Whats the radius of convergence of this power series. We also have to show that $\nu(z) = \sum\limits_{n=1}^{\infty} \frac{z^{n}}{1-z^{n}}$
Regarding the divisor function, we have Dirichlet's formula in hand. But will that help!