Convergence radios for taylor series for the function $f(z)=\frac{1}{1+e^z}$ at $z=0$

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 1

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 $\pi$.

Answer 2

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

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

Answer 3

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.

Answer 4

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.