It's well known that
$$\frac{1}{e^x-1}=\sum_{n=0}^{\infty}B_n\frac{x^{n-1}}{n!}\hspace{1cm}\left(\text{or}\hspace{0.5cm} \frac{x}{e^x-1}=\sum_{n=0}^{\infty}B_n\frac{x^{n}}{n!}\right)$$ where $B_n$ are the Bernoulli numbers.
I want tho obtain the series of this function but centered in a generic point $x_0=a \neq 0$, that is:
$$f(x)=\frac{1}{e^x-1}=\sum_{n=0}^{\infty}f^{(n)}(a)\frac{(x-a)^n}{n!}$$
and I wonder if it's possible write the n$-th$ derivative $f^{(n)}(a)$ in terms of Bernoulli numbers $B_n$
The result is directly related to the Eulerian polynomials $A_n(t)$.
This can be shown using the expression for $\sum_{n=0}^\infty A_n(t)z^n/n!$; another way is to use $$\sum_{k=1}^\infty k^n t^k=\frac{t A_n(t)}{(1-t)^{n+1}}\qquad(|t|<1)$$ as the definition of $A_n(t)$: assuming $\Re a>0$, for "small enough" $x$ we have $$\frac{1}{e^{a+x}-1}=\sum_{k=1}^\infty e^{-k(a+x)}=\sum_{k=1}^\infty e^{-ka}\sum_{n=0}^\infty\frac{(-kx)^n}{n!}=\sum_{n=0}^\infty\frac{(-x)^n}{n!}\frac{e^{-a}A_n(e^{-a})}{(1-e^{-a})^{n+1}},$$ and this implies the following (which extends to $a\notin 2\pi i\mathbb{Z}$ by analytic continuation): $$(-1)^n f^{(n)}(a)=\frac{e^{-a}A_n(e^{-a})}{(1-e^{-a})^{n+1}}\color{blue}{\underbrace{{}=\frac{e^a A_n(e^a)}{(e^a-1)^{n+1}}}_{n>0}}.$$
Multiplying the above series by $a+x$ (not by $x$!) and taking $a\to 0$, we recover the power series for $x/(e^x-1)$, expressing the Bernoulli numbers as a limit involving the Eulerian polynomials.