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?
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 $z=1$.