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 theorem that analytic functions are equal to their own Taylor series tells you that the Taylor series around a point $a$ converges on every open disk around $a$ where the function is analytic. Here, $f(z)=\frac1{e^z+1}$ is analytic when $e^z\neq -1$, i.e. when $z\neq i\pi +2\pi k i$, $k\in\mathbb{Z}$. 

You can find out the Taylor series of your function by plugging in the series for $e^z$. However, you need to be careful. One way that works is to write
$$\frac1{1+e^z} = \frac1{2+e^z-1}=\frac12 \frac1{1+\frac{e^z-1}{2}}
$$
and then to apply the series for $\frac1{1+w}$ to $w=\frac{e^{z}-1}{2}$. The point in using $\frac{e^{z}-1}{2}$ is that it is zero at your expansion point, which is a condition for plugging a series into the geometric series.