Convergence of series to analytic function Was going over some old complex analysis questions and came across this one that requires the ratio test I think:
$\displaystyle
\sum\limits_{n}{\frac{z^n}{1+z^n}}
$
and I can't quite seem to figure it out. We need, $$\left |\frac{\frac{z^{n+1}}{1+z^{n+1}}}{\frac{z^n}{1+z^n}} \right | <q< 1$$ which is true if and only if $$\left |\frac{z(1+z^n)}{1+z^{n+1}} \right|<q < 1$$ but it is not clear where to go from here. Any hints someone could provide?
 A: When $|z|>1$, the general term does not tend to zero.
When $|z|<1$, the general term is asymptotic to $z^n$.
When $|z|=1$, if $z^n=-1$ for some $n$ (i.e. $z$ is a rational power of $-1$), the series is not defined; else $z^n+1$ can be as small as you want and the series does not converge.
A: I think this is easier to do using dominated convergence directly instead of the ratio test. For $\vert z\vert<1$ we have $\left\vert\frac{z^n}{1+z^n}\right\vert=\frac{\vert z\vert^n}{\vert 1+z^n\vert}\leq\frac{\vert z\vert^n}{1-\vert z\vert^n}\leq\frac{\vert z\vert^n}{1-\vert z\vert}$. The first inequality follows from the inverse triangle inequality, and the second one from $\vert z\vert^n\leq\vert z\vert$ for all $\vert z\vert<1$. With this, your series is dominated by a geometric series, so it converges absolutely.
But the more interesting question is wether the limit is an analytic function. This is only guaranteed if the convergence is uniform on compact subsets of the unit disc. Luckily, this is the case, since on the compact disc $\overline U_r(0):=\{z\in\mathbb C~\vert~\vert z\vert\leq r<1\}$ your series is dominated by the series $\sum_n \frac{r^n}{1-r}$, so convergence on the compact discs is uniform due to the Weierstraß-M-test.
