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The following is an old qualifying exam question:

Let $f(z) = \sum_{n=0}^{\infty}a_nz^n$ have radius of convergence $R$. If $|a| = R$ and the power series does not converge at $z = a$, then $f(z)$ cannot be analytically continued to an open neighborhood of $a$.

So I'm thinking that this is not true: for instance, take $f(z) = \sum_{n=0}^\infty z^n$. This has radius of convergence $1$, the power series does not converge at 1, but nevertheless, $f(z) = \frac{1}{1-z}$ which is analytic everywhere except 1; so namely, on an open neighborhood of $1$. Do I have this right?

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  • $\begingroup$ Don't neighborhoods of a point have to include the point itself? $\endgroup$ – N8tron May 28 '18 at 2:40
  • $\begingroup$ @N8tron yep! whoops. $\endgroup$ – Möbius Dickus May 28 '18 at 2:43
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    $\begingroup$ The issue persists, considering $a=-1$ instead. I also suspect that the statement is not true. $\endgroup$ – Sangchul Lee May 28 '18 at 2:56
  • $\begingroup$ What about a function with a removable singularity at, say, 1, and a radius of convergence 1? $\endgroup$ – Möbius Dickus May 28 '18 at 2:57
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    $\begingroup$ Removable singularities are artificial in that they appear only by means of puncturing the domain, and as such, power series cannot detect removable singularities. Anyway, the statement would be true if it claims that there exists $a$ with $\lvert a\rvert=R$ such that $f$ cannot be analytically continued to an open neighborhood of $a$. $\endgroup$ – Sangchul Lee May 28 '18 at 3:04
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Not true in general, for instance $$1/(1-z)=\sum_{n\ge 0} z^n$$ has radius of convergence $1$, not convergent for $z=-1$, but can be analytically continued around $z=-1$. The same thing is true for any $z$, $|z|=1$, $z\ne 1$.

However, it cannot be analytically continued around $z=1$, and that is a general fact for series with positive coefficients.

Compare with $$\log\frac{1}{1-z}=\sum_{n\ge 1} \frac{z^n}{n}$$ with radius $1$. For every $z$ with $|z|=1$, $z\ne 1$, the series is convergent , and can be analytically extended around $z$. Again, it will not be convergent for $z=1$, and cannot be analytically extended around $1$.

Third example $$\sum_{n \ge 1}\frac{z^n}{n^2}$$ radius $1$, uniformly convergent on the closed disk of radius $1$, can be analytically continued around any point $z$, $|z|=1$, except $z=1$.

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