Closed form for Taylor series of $\frac{x}{\sinh(x)}$

Question

I'm trying to obtain a closed form for the coefficients of the Taylor expansion of the function $\frac{x}{\sinh(x)}$ near $0$.

I have calculated up to order 15 with a program and the expansion is $$ - \frac{8191 x^{14}}{37362124800} + \frac{1414477 x^{12}}{653837184000} - \frac{73 x^{10}}{3421440} + \frac{127 x^{8}}{604800} - \frac{31 x^{6}}{15120} + \frac{7 x^{4}}{360} - \frac{x^{2}}{6} + 1 $$ i didn't recognize these numbers at all. I know that such function is the inverse of the even part of the function $f(x)=\frac{1-e^x}{x}$. The coefficients of the function $f(x)^{-1}$ admits a closed form in terms of Bernoulli number, so i was hoping that this function admits a closed form, too. Did you know any reference where i can find tables of Taylor series?

Answer 1

Hint. One may notice that $$ \frac{x}{\sinh(x)}=\frac{2x}{e^x-e^{-x}}=\frac{2x}{1-e^{-x}}-\frac{2x}{1-e^{-2x}} $$ then use the standard generating function:

https://mathworld.wolfram.com/BernoulliNumber.html.

Answer 2

Use Mittag-Leffler's theorem, expansion at poles:

$$\frac{1}{\sinh z}=\frac{1}z+2z\sum_{n=1}^\infty \frac{(-1)^n}{z^2+n^2\pi^2}$$

Then we have:

$$\begin{align}\frac{z}{\sinh z}&=1+2z^2\sum_{n=1}^\infty \frac{(-1)^n}{z^2+n^2\pi^2}\\ \\ &=1+2z^2\sum_{n=1}^\infty \frac{(-1)^n}{n^2\pi^2}\cdot\frac{1}{1+\frac{z^2}{n^2\pi^2}}\\ \\ &=1+2z^2\sum_{n=1}^\infty \frac{(-1)^n}{n^2\pi^2}\cdot\sum_{k=0}^\infty(-1)^k\frac{z^{2k}}{n^{2k}\pi^{2k}}\\ \\ &=1+\frac{2}{\pi^2}z^2\sum_{k=0}^\infty\sum_{n=1}^\infty \frac{(-1)^n}{n^{2k+2}}\cdot(-1)^k\frac{z^{2k}}{\pi^{2k}}\\ \\ &=1+\sum_{k=0}^\infty(-1)^k\cdot\left(\frac{1}{4^{k}}-2\right)\cdot\frac{\zeta(2k+2)}{\pi^{2k+2}}\cdot z^{2k+2} \end{align}$$