Someone asked a question here about why the Taylor series of $\log(1+x)$ diverges:
Why does the taylor series of $\ln (1 + x)$ only approximate it for $-1<x \le 1$?
I have a similar question: why does the $\operatorname{Sech}(x)$ Taylor series diverge at $\pi/2$?
The answer in the $\log$ question dealt with the fact that the only symmetric interval that could be defined for $\log$ is from $-1$ to $1$ as $\log$ is only defined from $(-1, \infty)$. $\operatorname{Sech}(x)$ is defined everywhere and it doesn't have this problem.
Thanks for all your help.
Easiest reason is in complex analysis. $\mathrm{sech}(x)$ has a pole at $i\pi/2$, so the radius of convergenge cannot extend beyond that point. And moreover, there are no singularities strictly closer to $0$, so the radius of convergence is exactly $\pi/2$.