$\frac{2x}{e^{2x}-1} = -2x(1-e^{2x})^{-1}$
We can obtain the binomial series expansion of $(1-e^{2x})^{-1}$:
$(1-e^{2x})^{-1} = \sum_0^\infty\begin{pmatrix}-1\\n\end{pmatrix}(-e^{2x})^{n} = \sum_0^\infty(e^{2x})^{n} = 1 + e^{2x} + e^{4x} + e^{6x}... $
Now, each of the expressions $e^{2x}, e^{4x}, e^{6x}, ...$ can itself be expanded using the Maclaurin series for an exponential function, i.e. $e^{x}=\sum_0^\infty\frac{x^{n}}{n!}$. Thus, we have
$\sum_0^\infty(e^{2x})^{n} = 1 + e^{2x} + e^{4x} + e^{6x}... = 1 + \sum_0^\infty\frac{(2x)^{n}}{n!} + \sum_0^\infty\frac{(4x)^{n}}{n!} + \sum_0^\infty\frac{(6x)^{n}}{n!} + ...= 1 + \sum_0^\infty[\frac{2^{n}x^{n}}{n!} + \frac{4^{n}x^{n}}{n!} + \frac{6^{n}x^{n}}{n!} + ...]$
And,
$1+\sum_0^\infty[\frac{2^{n}x^{n}}{n!} + \frac{4^{n}x^{n}}{n!} + \frac{6^{n}x^{n}}{n!} + ...] = 1 + \sum_0^\infty[\frac{2^{n}x^{n}}{n!}(1 + 2^{n} + 3^{n} + ...)] = 1 + \sum\limits_{n=0}^\infty\frac{2^{n}x^{n}}{n!}\sum\limits_{k=1}^\infty k^{n}$
Next, we can multiply the last expression by $-2x$ to obtain the required series. But that turns out to be wrong, for if I expand the series thus obtained, I get something completely different from the correct expansion, which is
$1 − x + x^{2}/3 − x^{4}/45 ...$
Where am I going wrong?
This is wrong because the expansion $(1-y)^{-1}=1+y+y^2+...$ is legal only when $|y|\lt 1$. For $(1-e^{2x})=1+e^{2x}+e^{4x}+...$ to be true, we must have $|e^{2x}|\lt 1$, which is not true for $x\gt 0$.