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?
 A: This is indeed true. Look for the poles of $f(z)$. Quite clearly, they occur when $e^{z} = -1$. The smallest solutions - in terms of argument - to this equation are $\pm i \pi$.
Hence, we see the closest poles of $f$ to the center of the desired Taylor are at a distance of $\pi$ in the complex plane. A standard result shows the radius of convergence of a Taylor series is equal to the distance of the center of the expansion to the nearest singular point (in this case a pole) of the function we seek to represent.
A: The function is analytic in $\{z:|z|<\pi\}$ so the power series expansion is valid in this disk. The radius of convergence is determined by  the nearest pole to $0$ and the nearest pole in this case is $z=i\pi$ so the radius of convergence is indeed $\pi$.
A: 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 $\pi$.
