Is $\pi$ the radius of convergence of the taylor series of $f(z)=\frac{1}{1+e^z}$? I'm having trouble with the following exercise:

is it true that the radius of convergence of the Taylor series of the function $$f(z)=\frac{1}{1+e^z}$$ centered at $z=0$ is $\pi$?

I tried to evaluate the taylor series combining the two identities: $$e^z=\sum_{n\geq0}\frac{z^n}{n!}$$
$$\frac{1}{1-z} = \sum_{n\geq0}z^n,|z|<1$$
but it got really messy very quickly and I wasn't able to conclude anything about the radius of convergence.
Given that this is a True/False type question, is it possible to conclude anything without actually evaluating the Taylor series? If so, how?

Intuitively the radius of convergence is the radius of the largest disk around $0$ that you can find in which $1/(1+e^z)$ is analytic, that's the distance to the closest points where $1 + e^z = 0$.  These points are $\pm i\pi$, so the radius of convergence is $\pi$.
