Let $f(z)=\sum_{n=0}^{\infty}\frac{z^{n}}{1-z^{n}}$. Then $f$ converges in the unit disc. I want to show that it is analytic in this region as well, but since it is not a power series I don't have any theorems to apply. My only thought was to try and use similar reasoning as with power series: show that converges is uniform so that we can switch integral and sum to show that the integral of the series is 0, then apply Morera's Theorem. But the problem is, according to Wolfram Alpha, $\int_{\gamma}\frac{z^{n}}{1-z^{n}}=\int_{0}^{2\pi}\frac{e^{int}ie^{it}}{1-e^{int}}dt$ doesn't converge.
EDIT: To apply Morera's Theorem $\gamma$ should be closed path contained in the unit circle. (It probably should have been obvious that the integral doesn't converge since the unit circle contains 1…). So the above strategy does works (yay!) since the function is holomorphic in any closed disk with radius less than 1.
Best Answer
Let $C$ be a compact subset of the open unit disk and let $M=\sup_{z\in C}|z|$. Then $M<1$ and, for each $n\in\mathbb N$,$$\left|\frac{z^n}{1-z^n}\right|=\frac{|z|^n}{|1-z^n|}\leqslant\frac{M^n}{1-|z|^n}\leqslant\frac{M^n}{1-M^n}.$$Since the series $\sum_{n=1}^\infty\frac{M^n}{1-M^n}$ converges, your series converges uniformly on $C$. Since this holds for every compact subset of $D_1(0)$, the sum of the series is analytic.