How fast do the MacLaurin coefficients of $\tanh(z)$ converge for $z = 1$? Does distance to nearest pole determine convergence rate?

Question

In the Oct 12 issues of SIAM News it states that for a mathematician not to know that the coefficients of the Taylor series for $\tanh(z)$ centered at $z=1$ decay as $(2/\pi)^n$ is like a physicist not to be able to work out when a stone dropped from a height of 10m hits the ground. I am ashamed to admit I am not familiar with this computation. I worked out a few terms of the Taylor series, but they're fairly messy, not obviously geometric. The article makes it sound like it's based on proximity to the pole at $z=i\pi/2$. Wikipedia makes no mention of this. Is there some relationship between the rate of convergence of a Taylor series and the proximity of the center to a pole?

Answer 1

The article does not speak of the Taylor series centered on $z=1$. It speaks of the Taylor (MacLaurin) series centered on $z=0$, evaluated at $z=1$.

The sketch would be:

1. $\tanh(z)$ has poles at the zeros of $\cosh(z)$, which are $z= \pi k i + \frac{\pi i}{2}$.

2. Hence the nearest poles to the origin are $\pm \frac{\pi i}{2}$

3. Hence the MacLaurin series $$ a_0 + a_1 z + a_2 z^2 + \cdots$$ converges on a circle that extends up to that point(s) (radius$= \pi/2>1)$.

4. Hence the MacLaurin series at $z=1$ (which concides with the series of the Maclaurin coefficients $ a_0 + a_1 + a_2 + \cdots $) converges.

5. Then, by Cauchy-Hadamard theorem $|a_n| \to R^{-n} = (2 /\pi)^n$ (in the article: "so So the Maclaurin coefficients decay at the rate $(2 /\pi)^n$")

6. Then the series $ a_0 + a_1 + a_2 + \cdots $ converges as a geometric series, at a rate $(2 /\pi)^n$.