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.
Best Answer
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.