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The series $\displaystyle\sum \dfrac{z^n}{n^2}$ converges for $\lvert z\rvert<1$ by the ratio test, meaning that the dilogarithm function $\text{Li}_2(z),$ which is equal to the series $\displaystyle\sum \dfrac{z^n}{n^2}$ when it converges, is certainly analytic on $\lvert z\rvert<1$.

Similarly, by the ratio test, the series diverges for $\lvert z\rvert>1$. And we know that for a meromorphic function, the radius of convergence is always the distance from the center to the nearest singularity(wikipedia). Hence we should conclude that this function has a pole somewhere on $\lvert z\rvert=1$?

On the other hand, at the point $z=1$ it converges to $\pi^2/6$ (see Basel problem), it must also converge at the point $z=-1$, and it is stated on this question by Ben that the series is convergent on the whole circle $\lvert z\rvert=1$ (this confused me and prompted the question), and on a comment to an answer to a related question, one is reminded by user 23rd that if a function is analytic on a closed disk, then it is analytic on an open disk of larger radius. So we should conclude that this series is convergent for some $z$ with $\lvert z\rvert>1$? For the whole complex plane? In fact, this Wolfram alpha page does claim that the function is analytic on all of $\mathbb{C}$ (if I'm reading it correctly; it's very terse).

Actually that second related question (singularity of analytic continuation of $f(z) = \sum_{n=1}^\infty \frac{z^n}{n^2}$) already contains the answer to my question: the dilogarithm is analytic on $\mathbb{C}\setminus [1,\infty)$. But I can't understand how that answer is consistent with the other remarks, and the Wolfram page. How is this situation reconciled? Could I get an explanation that's a little more detailed than what's already there?

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  • $\begingroup$ " if a function is analytic on a closed disk, then it is analytic on an open disk of larger radius" But we don't know this series is analytic on $\{|z| \le 1\}.$ $\endgroup$
    – zhw.
    Commented Nov 4, 2015 at 19:11
  • $\begingroup$ @zhw.: doesn't convergence at a point imply analyticity? $\endgroup$
    – ziggurism
    Commented Nov 4, 2015 at 19:59
  • $\begingroup$ No. In fact $(\sum x^n/n^2 - \pi^2/6)/(x-1) \to \infty$ as $x\to 1^-.$ $\endgroup$
    – zhw.
    Commented Nov 4, 2015 at 20:09
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    $\begingroup$ @zhw.: right. It is convergent in an open neighborhood of a point which is equivalent to differentiable. We have convergence at $z=1$, but no neighborhood thereof $\endgroup$
    – ziggurism
    Commented Nov 4, 2015 at 20:52
  • $\begingroup$ By the way, the dilogarithm function isn't meromorphic, because of the branch point and branch cut. $\endgroup$
    – tparker
    Commented Sep 17, 2018 at 17:37

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The alternative definition (easily proven equivalent to the one you give for $\lvert z \rvert < 1$) $$ \operatorname{Li}_2 (z) = \int_0^z \frac{\log{(1-t)}}{t} \, dt, $$ where the integral can be taken to be along the ray joining $0$ to $z$ for $z \notin (1,\infty) $, shows that the dilogarithm has a branch point at $z=1$ (since the logarithm inside the integral has one): you can see that no extension across the cut is possible by computing the derivative across the cut, for example.

As we see here, it is possible for a power series to converge on its whole circle of convergence (since the series is absolutely convergent for $\lvert z \rvert \leqslant 1$), but for no larger $\lvert z \rvert$; in this case, because the branch cut "gets in the way" of defining an analytic extension to a larger convex domain.

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  • $\begingroup$ So a statement like "the radius of convergence is always the distance to the nearest pole" is not correct and this function is a counterexample because there is no pole at $z=1$. It should be "radius of convergence is distance to nearest pole or branch point"? That's what the linked Wikipedia article means by "singularity"? $\endgroup$
    – ziggurism
    Commented Nov 4, 2015 at 21:15
  • $\begingroup$ Yes, although the terminology is likely not universal. Probably the best way to think about a singularity is a point $a$ such that there is no open disk containing $a$ on which the function is analytic. This includes poles, essential singularities and branch points, which are essentially the only possible ways to lose analyticity at a point (if the point is isolated, it's one of the first two, else it's a branch point). $\endgroup$
    – Chappers
    Commented Nov 5, 2015 at 1:40
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    $\begingroup$ I guess this question has little to do with special functions like the dilogarithm. If you want a series to exhibit the phenomenon that a branch point can end the radius of convergence instead of a pole, the series for $\sqrt{1-z}$ behaves in exactly the same way, though I found that series harder to work with. $\endgroup$
    – ziggurism
    Commented Nov 5, 2015 at 4:27
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    $\begingroup$ @ziggurism Actually isolated singularities and branch points is rather an exceptional way to lose analyticity. The generic one would correspond to emergence of natural boundaries, such as the circle $|z|=1$ for the series $\sum_{n}z^{n^2}$. $\endgroup$
    – Start wearing purple
    Commented Nov 11, 2015 at 11:53
  • $\begingroup$ @Startwearingpurple: can you elaborate on that idea? What is a natural boundary? Boundary between convergent and divergent domains? Are there some kind of singularities on this boundary? I don't recognize the series you quoted $\endgroup$
    – ziggurism
    Commented Nov 13, 2015 at 2:05

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