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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?

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    Now I understand why z=1 is a singularity, if you differentiate this series term by term, you get harmonic series.2012-10-15

2 Answers 2

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This series defines the dilogarithm (see too here) and may be rewritten as (expand the $\log$ in series and integrate to get this) : $\operatorname{Li}_2(z)=\sum_{v=1}^\infty \frac {z^v}{v^2}=-\int_0^z \frac {\log(1-t)}t\,dt$

The function $\operatorname{Li}_2$ does not have poles nor essential singularities but two branch points at $z=1$ and $z=\infty$ (see too the $\Im(\operatorname{Li}_2(z))$ picture from the first link). The $z=1$ branch point should be your answer.

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Since the coefficients are all positive, a theorem of Pringsheim says there is a singularity at $z=1$.

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    I knew this theorem, I just can't see why2012-10-15