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Given a power series $\sum_{k=0}^\infty a_k z^k$ with radius of convergence $0<R<\infty$ . Given a point in the boundary of the circle $z, |z|=R$, is there a relationship between $z$ being a regular point and the convergence of the series at $z$ ? (By regular point I mean that the series has a direct analytic continuation with a circle with center $z$)

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Whether the series converges or not, the point can be either regular or not regular.

An example when the series converges and the point is regular: $\log (1+z)$ at $z=1$: $$1-\frac12+\frac13-\frac14+\dots=\log 2.$$ An example where the series does not converge but the point is regular: $1/(1+z)$ at $z=1$: $$1-1+1-1+\dots \ \ \ \rm does\ not\ converge.$$ An example where the series does not converge and the point is not regular: $1/(1+z)$ at $z=-1$: $$1+1+1+1+\dots \ \ \ \rm does\ not\ converge.$$ An example where the series does converge but the point is not regular: $$f(z):=\sum_{n\ge 1} \frac{z^{n^2}}{n^2} \qquad {\rm at} \qquad z=1:$$ $$1+\frac14+\frac19+\dots \rm = \frac{\pi^2}{6}$$ but $$\lim_{r\to 1^-} f'(r)=\lim_{r\to 1^-} \sum_{n\ge 1} r^{n^2-1} = +\infty $$ so $f$ has a singularity at $1$.

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I don't believe there is an implication in either direction.

It's easy to see that $z$ being a regular point doesn't imply that the series converges at $z$: take $f(z) = 1/(1-z) = \sum_{k=0}^\infty z^k$. Every point on the unit circle other than $z=1$ is a regular point, but the series doesn't converge anywhere on the unit circle (the terms don't tend to $0$).

I think the following is a construction of a series with radius of convergence $1$ that converges at $z=1$ but $z=1$ is not a regular point. Let $$ f(z) = \sum_{n=1}^\infty \frac1{n!} \frac1{1-z/(1+1/n)} = \sum_{n=1}^\infty \frac1{n!} \sum_{k=0}^\infty \frac{z^k}{(1+1/n)^k} = \sum_{k=0}^\infty z^k \sum_{n=1}^\infty \frac1{n!(1+1/n)^k}. $$ It's late and I'm not thinking clearly ... maybe some commenters can justify the exchange of summation, see why the radius of convergence equals 1 and why the series converges at $z=1$. The point though is that $f(z)$ has poles at the points $2,3/2,4/3,5/4,\dots$ and so cannot be analytically continued in any neighborhood of $z=1$.

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