Take a function that is analytic at 0 and consider its Maclaurin Series. Here are some examples I'll refer to:

$$\frac{1}{1-x} =\sum_{n=0}^\infty x^n$$ $$\frac{1}{1+x^2} =\sum_{n=0}^\infty(-1)^nx^{2n}$$ $$\ln(1-x) =-\sum_{n=1}^\infty\frac{x^n}{n}$$ $$\sqrt{1-x} =1-\sum_{n=1}^\infty\frac{(2n-2)!}{2^{2n-1}n!(n-1)!}x^n$$

Each of these series has a radius of convergence of 1. And each function either

$\bullet$ has a singularity along the edge of the disk of convergence (at 1, $\pm i$, and 1 in the first three examples respectively) or

$\bullet$ has a derivative with a singularity along the edge of the disk of convergence (the last example is this way at 1).

My question is: Suppose a function $f$ is analytic at 0 and its Maclaurin Series has a radius of convergence $r<\infty$. Does it have to be the case that some derivative (0th, 1st, 2nd, ...) of $f$ blows up somewhere along the edge of the disk of convergence?


$f(x)=\sum x^n/e^{\sqrt n}$ has radius of convergence 1, and it and all its term-by-term derivatives converge everywhere on the unit circle. Basically, $e^{\sqrt n}$ goes to infinity faster than any polynomial but more slowly than any exponential.

  • $\begingroup$ Thanks! That's just the kind of example I was looking for. $\endgroup$ – alex.jordan Jun 24 '11 at 6:27
  • $\begingroup$ Interesting series - if you consider it a Laurent series, it actually also has a lower convergence radius of one, i.e. it only converges for $|x|=1$ if $x\in\mathbb C$ is allowed... It's probably due to the imaginary part jumping everywhere else for $\Im x\neq0$ $\endgroup$ – Tobias Kienzler Jul 15 '13 at 13:20
  • $\begingroup$ @TobiasKienzler: What do you mean by "it only converges for $|x|=1$"? Are you claiming it diverges for $|x|<1$? If so, that is false. $\endgroup$ – Hans Dec 27 '18 at 3:16
  • $\begingroup$ @Hans It's been a few years since I posted this, so I may misremember, but: If the series is generalized to $$f(x) = \sum_{n=-\infty}^\infty x^n / e^{\sqrt{|n|}}$$ the inner radius of convergence becomes $r = \limsup_{n\to\infty} |e^{-\sqrt{|n|}}|^{\frac1n} = 1$ as well. My link was wrong then though, since the negative $n$ part was missing, here's a fix, see also this image. $\endgroup$ – Tobias Kienzler Dec 27 '18 at 16:15
  • $\begingroup$ @TobiasKienzler: Now this reformulation makes sense. In fact, it makes sense for not taking the absolute value of $n$ as well. $\endgroup$ – Hans Dec 27 '18 at 20:46

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