the first $2k$ terms of the power series of $\sec x + \tan x$ at $x=-\pi/2$

Question

We know the power series of $\sec x+\tan x$ is as follows,

$f(x)=\sum_{n\geq 0}\frac{E_n}{n!}x^n$, where $E_n$ is Euler Zigzag numbers and clearly the radius of convergence of $f(x)$ is $\pi/2$.

Suppose $f_k=\sum_{n\geq 0}^{2k-1}\frac{E_n}{n!}x^n$, i.e., the sum of the first $2k$ terms.

Obviously $f(x)$ diverges at $-\frac\pi2$. However it seems $\lim_{k\rightarrow \infty}f_k(-\frac{\pi}2)$ converges.

I calculated first several terms of $f(x)$ and found $f_k(-\frac\pi 2)$ seems to converge to $-\frac2\pi$.

But I have no idea to prove it. Can someone give me some hints?

Thanks in advance!

Answer 1

By Abel's Theorem, if the power series converges, then it agrees with the limit of the function.

This argues against $-2/\pi$. Instead, either the power series converges to $0$ (by the limit Michael & Ewan give in the other answer) or it does not converge.

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Some messing around numerically: It looks like the terms of the sum (not the sum itself) approach $(-1)^n 4/\pi$.

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Computation of the sum of first $2k$ terms, as $k \rightarrow \infty$:

As Winther said, we're adding up $$\left(\frac{E_{2n}}{(2n)!} - \pi/2 \frac{E_{2n+1}}{(2n+1)!}\right)\left(\pi/2\right)^{2n}$$

This is exactly the Maclaurin series for $g(x) = \mathrm{sec}(x) - \frac{\pi}{2} \mathrm{tan}(x)/x$ evaluated at $x = \frac{\pi}{2}$.

We need: 1. Convergence of the series at $x = \frac{\pi}{2}$. And 2. The limit of the function to be as desired.

Here's 2:

$\mathrm{lim}_{x \rightarrow \frac{\pi}{2}} \mathrm{sec}(x) - \frac{\pi}{2} \mathrm{tan}(x)/x = \mathrm{lim}_{x \rightarrow \frac{\pi}{2}} \frac{x - \frac{\pi}{2} \mathrm{sin}(x)}{x\, \mathrm{cos}(x)} = \mathrm{lim}_{x \rightarrow \frac{\pi}{2}} \frac{1-\frac{\pi}{2}\mathrm{cos}(x)}{\mathrm{cos}(x) - x\, \mathrm{sin}(x)} = \frac{1}{-\pi/2} = -\frac{2}{\pi}$.

For 1, notice $g(z)$ is holomorphic out past $\pi/2$ (only possible poles are at $\pm \pi/2$ and the limit for $-\pi/2$ is $2/\pi$ and we already computed the limit at $+\pi/2$).

Answer 2

Consider the limit

$$\lim_{n\to \infty} \frac{E_n}{n!}\left(\frac{\pi}{2}\right)^n$$

Using $E_{n} \sim \frac{(-1)^{(n-1)/2} 4^{n+1}}{n+1}B_{n+1}$ together with $B_{2n} \sim (-1)^{n+1}4\sqrt{\pi n}\left(\frac{n}{\pi e}\right)^{2n}$ and Stirlings approximation we get

$$\lim_{n\to \infty} \frac{E_{2n-1}}{(2n-1)!}\left(\frac{\pi}{2}\right)^{2n-1} = \frac{4}{\pi}$$

so the terms in the series does not converge to $0$ and consequently the series does not converge.