I need to show that $\sum_{n=1}^{\infty} \frac{z^n}{n^2}$ is continuous in the closed unit disk and holomorphic in the open unit disk but not holomorphic in the the closed disk.
I showed that the series is convergent absolutely and uniformly on compact subsets of the closed unit disk (by the M-test), then the series is holomorphic (and so continuous ) but still not sure how to prove that it is continuous on the boundary not holomorphic on the boundary.
On
This is the di-logarithm function. It has radius of convergence $1$, by Cauchy-Hadamard. It converges absolutely on the boundary, by comparison with $\sum_{n\ge0}1/n^2$.
It's derivative is $g (z)/z $, where $g (z)=-\ln (1-z) $. $g (z)=\sum_{n\ge0}z^n/n $ doesn't converge at $z=1$. Thus it is not holomorphic on the closed disk.
The series converges uniformly on the closed unit disk, by the $M$-test, so it is continuous there. To be holomorphic at a point, a function must be differentiable in a neighborhood of the point. If it is holomorphic on the closed disk, then for every $z$ with $|z|=1$ there is an open disk centered at $z$ on which the function is holomorphic. Since the boundary is compact, finitely many such disks cover the boundary, so the series is holomorphic is some disk $|z|<r$ where $r>1$. This is absurd, since the radius of convergence is $1$.
Actually, we know by Pringsheim's theorem that the the function cannot be holomorphic in a neighborhood of $1$.