Question: I'm trying to find the Convergence radios for the function $f(z)=\frac{1}{1+e^z}$ using taylor series around $z=0$. this is what I've got so far:

$\frac{1}{1+z} = \sum\_{n=0}^{\infty}z^n$ thus I get:

$\frac{1}{1+e^z}=\sum\_{n=0}^{\infty}(\sum\_{m=0}^{\infty}\frac{n^mz^m}{m!})$

also I'm having a problem because $z$ should be to the power of $n$ not $m$.

now I can use $\frac{1}{R}=limsup\_{n->\infty}(a\_n)^{\frac{1}{n}}$ but $a\_n=\sum\_{m=0}^{\infty}\frac{n^m}{m!}$ and I don't know how to calculate $limsup\_{n->\infty}(\sum\_{m=0}^{\infty}\frac{n^m}{m!})^\frac{1}{n}$. please help!

by the way, could I have first used the taylor series for $e^z$ and than plug it in $\frac{1}{1+e^z}$? I got stuck there too but I'm wondering if it's even possible.

Answer: 
The radius of convergence is $\pi$. It is the radius of the largest disc around $0$ in which the function is analytic. By solving $e^{z}+1=0$ you can see that the radius is $\pi$.
The radius is $\pi$. In fact, for every analytic function $g$ and every $z\_0\in D\_g$, if $r>0$ is such that $D(z\_0,r)\subset D\_g$, then the radius of convergence of the Taylor series of $g$ centered at $z\_0$ is at least $r$.

In your case, the largest domain that $f$ can have is $\mathbb{C}\setminus\{\pi i+2\pi i n\,|\,n\in\mathbb Z\}$. So, the largest disk centered at $0$ contained in $D\_f$ is $D(0,\pi)$ and therefore the radius of convergence is at least $\pi$.

But it cannot be largest than that. Otherwise, if $\sum\_{n=0}^\infty a\_nz^n$ is the Taylor series, we would have\begin{align}\sum\_{n=0}^\infty a\_n(\pi i)^n&=\lim\_{z\to\pi i}\sum\_{n=0}^\infty a\_nz^n\\&=\lim\_{z\to\pi i}\frac1{1+e^z},\end{align}but this limit doesn't exist.