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 isolated singularities of $f$ are:

$$ z\_k:=(2k+1) i \pi, \quad k \in \mathbb Z.$$

hence the radius in question is given by 

$$ min \{|z\_k|:k \in \mathbb Z\}= \pi.$$