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So the power series $\sum\limits_{n \geq 0} z^n $ converges in the unit disc and is holomorphic in the unit disc as well with the derivative $\sum\limits_{n \geq 0} nz^{n-1}$. I am trying to prove that the power series above cannot be continued analytically past the unit disc.

What I was thinking is that I can say that the derivative needs to be continuous, so if the series had an extension past the unit disc, the derivative on a point of the boundary of the unit disc would need to be $\lim\limits_{|z|\to 1^-} \sum\limits_{n \geq 0} nz^{n-1} $, which diverges since $|nz^{n-1}| \to \infty$. So the power series cannot be continued analytically.

I'm getting the feeling that such an argumentation is not complete or even incorrect. Any input is appreciated.

On the other hand, assuming that the argumentation above is okay, it seems to be not applicable on the case $\sum\limits_{n \geq 0} z^{2^n} $

mikasa
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    If you define $f\colon\Bbb C\setminus{1}\longrightarrow\Bbb C$ by $f(z)=\frac1{1-z}$, then $f$ is analytic, its domain strictly contains the open unit disk $D_1(0)$ and $(\forall z\in D_1(0)):f(z)=\sum_{n=0}^\infty z^n$. – José Carlos Santos Oct 29 '23 at 13:46
  • Okay, but what part of my argumentation was incomplete/wrong? @JoséCarlosSantos – mikasa Oct 29 '23 at 13:54
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    This function has an analytic continuation outside the unit disk but is not holomorphic at the point $z=1$. Namely you need to argue that as $z\rightarrow 1$ $f(z) \rightarrow \infty$ so any extension must have a pole there – Sidharth Ghoshal Oct 29 '23 at 13:59
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    For the other points on the boundary, at least of the form $M = e^{i\pi \frac{p}{q}}$ where $p,q$ are coprime integers it can be shown that the series $M^n$ is periodic and sums to 0 (for example at $M=-1$, $f(M) = 1-1+1-1…$, $1-1=0$) so you can’t be sure the function isn’t holomorphically continuable there (in fact it is continuable everywhere except $z=1$) – Sidharth Ghoshal Oct 29 '23 at 14:02
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    Analytic continuation and convergence of power series are very different issues in general - convergence of power series is given by the radius of convergence hence the coefficients, while analytic continuation is much subtler and has little to do with coefficient size in general – Conrad Oct 29 '23 at 14:02
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    Among other things, why would matter that $\lim_{|z|\to1^-}\sum_{n\geqslant0}nz^{n-1}$ diverges? Actually, for each $\omega$ such that $|\omega|=1$, then, unless $\omega=1$, the limit $\lim_{z\to\omega}\sum_{n\geqslant0}nz^{n-1}$ does exist; it is equal to $\frac1{(1-\omega)^2}$. – José Carlos Santos Oct 29 '23 at 14:06
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    This is implicit in existing comments, but to be explicit, with $f(z) = \sum_n z^n$ denoting the sum of the geometric series, it sounds as if you're treating the following conditions as equivalent when they are not: 1. $f$ has an analytic continuation beyond the unit disk. (True.) 2. The geometric series (or some other power series) converges to $f(z)$ on a disk of radius greater than $1$ (i.e., $f$ has an analytic continuation to a disk centered at $0$ and of radius greater than $1$). (False.) – Andrew D. Hwang Oct 29 '23 at 16:23

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