The radius of convergence of the series $\sum_{v\ge 1}v^{-2}z^v$ is 1, so there must be a singular point on the boundary. But for every $|\zeta|=1$,
$|\sum_{v\ge 1}v^{-2}\zeta^v|\le\sum_{v\ge 1}v^{-2}|\zeta|^v=\frac{\pi^2}{6}$
So this series is convergent on the boundary, but where is the singular point? Or am I making some mistakes?