Let $$ f(x) = \frac{e^{2\pi x}}{(1 - e^{2 \pi x})^2} - \frac{1}{(2\pi x)^2}$$ Can someone provide a formula for
$$ \lim_{x \to 0} \frac{d^k f(x)}{dx^k}$$ or at least for $\frac{d^k f(x)}{dx^k}$ ? Everything I've tried so far was L'Hospital but after few derivatives I've gave up and used Wolfram Alpha to test my results ... I've also been thinking about using the Cauchy Integral formula since $f(z)$ is holomorphic (I think) on $\mathbb{C}\setminus \{ i\cdot n| n\in \mathbb{Z}\}$ and can be continued analytically to $\mathbb{C}$ using Riemann theorem on removable singularities ... Can someone help?
Bernoulli numbers are usually defined through $$ \frac{z}{e^z-1} = \sum_{n\geq 0}\frac{B_n}{n!} z^n \tag{1}$$ Now we substitute $z\mapsto 2\pi z$, leading to: $$ \frac{2\pi}{e^{2\pi z}-1}=\frac{1}{z}+\sum_{n\geq 1}\frac{B_n(2\pi)^n}{n!} z^{n-1} \tag{2} $$ then differentiate both sides with respect to $z$, leading to: $$ -\frac{4\pi^2 e^{2\pi z}}{(e^{2\pi z}-1)^2}=-\frac{1}{z^2}+\sum_{n\geq 2}\frac{B_n(2\pi)^n (n-1)}{n!} z^{n-2} \tag{3}$$ By rearranging it follows that: $$ \frac{e^{2\pi z}}{(e^{2\pi z}-1)^2}-\frac{1}{4\pi^2 z^2}=-\sum_{m\geq 0}\frac{B_{m+2}(2\pi)^{m}}{(m+2)m!} z^{m} \tag{4}$$ and: $$ \frac{d^m}{dz^m}\left.\left(\frac{e^{2\pi z}}{(e^{2\pi z}-1)^2}-\frac{1}{4\pi^2 z^2}\right)\right|_{z=0} = \color{red}{-\frac{B_{m+2}(2\pi)^m}{(m+2)}}.\tag{5}$$