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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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"so there must be a singular point on the boundary" Why? –  user17762 Oct 15 '12 at 21:17
    
If a function is regular at a point, then so is its derivative. Will that help? Once you find the "bad" point, try expanding in a series centered there to find what kind of singularity it is. As you note, it is NOT a pole. –  GEdgar Oct 15 '12 at 21:18
    
@Marvis There is a theorem states that on the boundary of the disc of convergence of a power series, there is always at least one singular point. It can be proved by reductio ad absurdum, generally, if there is no singular point on the boudary, we can increase the radius of convergence. See Reinhold Remmer's Theory of Complex Functions P.234 for reference. –  Urukec Oct 15 '12 at 21:24
    
Now I understand why z=1 is a singularity, if you differentiate this series term by term, you get harmonic series. –  Urukec Oct 15 '12 at 22:52
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2 Answers

up vote 2 down vote accepted

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 why –  Urukec Oct 15 '12 at 22:34
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