# Why is $\operatorname{Sech}(x)$ Taylor series divergent past $\pi/2$?

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.

• The reason is the same as for the Taylor series of $1/(1+x^2)$ that only converges for $-1<x<1$. Commented Jul 12, 2016 at 16:02
• Radius of convergence is answered. But how about convergence at the point $\pi/2$, which is on the boundary of the region of convergence? Commented Jul 12, 2016 at 16:16
• the major contribution of comples analysis is that there is an equivalence between "a power series, analytic on a disk" and "no singularity on that disk", i.e. $f(z) = \sum_{n=0}^\infty c_n z^n$ always has a singularity on the boundary of its disk of convergence, and conversely, it doesn't converge past its largest singularity-free disk Commented Jul 12, 2016 at 16:46
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 $\pi/2$.