I'm wondering if there is a closed form expression for $$ \sum_{k=0}^\infty \frac{{\rm B}_{k+1}}{(k+1)!} \, \frac{(k-s)!}{k!(-s)!} \, \left(-x\right)^k $$ where ${\rm B}_k$ are the Bernoulli numbers and ${\rm Re}(s)>1$.
The expression has radius of convergence $2\pi$. I was thinking maybe it could be converted into a known series by methods of borel-transform or the like, but I haven't been able to figure it out :-(
If there is no closed form expression: I'm actually only interested in the large $x$ limit (which here is understood as analytic continuation), so maybe that might be simpler.
Any suggestions, also for certain $s$ values? Are there methods to extract the large $x$ limit?
edit: I guess I found some way to tackle the sum. Since it's only one term, I let the sum start at $k=1$ instead of $0$. The function is also odd, so writing $x$ instead of $-x$ too.
Now replacing \begin{align} (k-s)! &= \int_0^\infty t_1^{k-s} \, e^{-t_1} \, {\rm d}t_1 \tag{1} \\ \frac{1}{k!} &= \frac{1}{2\pi i} \int_{-\infty}^{0^+} e^{t_2} \, t_2^{-k-1} \, {\rm d}t_2 \tag{2} \end{align} where in the last integral the path is starting at $-\infty$ encircling $0$ once in positive direction and going back to $-\infty$ (Hankel contour).
Therefore without writing the integrals, unnecessary constants and shorthand $z=\frac{x t_1}{t_2}$ \begin{align} \sum_{k=1}^\infty \frac{{\rm B}_{k+1}}{(k+1)!} \, z^k = \frac{\coth\left(\frac{z}{2}\right)}{2} - \frac{1}{z} = \sum_{k=1}^\infty \frac{2z}{z^2 + \left(2\pi k\right)^2} = \sum_{k=1}^\infty \frac{2xt_1t_2}{\left(xt_1\right)^2 + \left(2\pi k \, t_2\right)^2} \, . \end{align} Next we multiply by $\frac{e^{t_2}}{t_2}$ and do the $t_2$-integral. In (2) before changing summation and integral order we are free to deform the contour as long as the only singularity at $0$ is enclosed. So shifting to a big circle $\gamma$ enclosing the "entire" complex plane we pick up the residues at $$t_2=\pm \frac{ixt_1}{2\pi k}$$ and arrive at the sum $$ \sum_{k=1}^\infty \frac{\sin\left(\frac{x t_1}{2\pi k}\right)}{\pi k} $$ which is why I asked the question about closed form here: Closed form expression for $\sum_{k=1}^\infty \frac{\sin\frac{x}{2\pi k}}{k \pi}$
While this one seems to be intractable I will also do the $t_1$-integration termwise. So multiplying by $t_1^{-s} \, e^{-t_1}$ and integrating while ${\rm Re}(s)<2$ $$ \int_{0}^\infty t_1^{-s} \, e^{-t_1} \, \frac{\sin\left(\frac{xt_1}{2\pi k}\right)}{\pi k} \, {\rm d}t_1 = {i\left(2\pi k\right)^{-s} \, (-s)!} \left\{ \left(2\pi k + ix\right)^{s-1} - \left(2\pi k - ix\right)^{s-1} \right\} $$ which can be continued to arbitrary $s$.
For the record we now have $$ f(x) = \sum_{k=1}^\infty \frac{{\rm B}_{k+1}}{(k+1)!} \, \frac{(k-s)!}{k!(-s)!} \, x^k = \sum_{k=1}^\infty {i\left(2\pi k\right)^{-s}} \left\{ \left(2\pi k + ix\right)^{s-1} - \left(2\pi k - ix\right)^{s-1} \right\} $$ where $(-s)!$ cancels.
I was ultimately only interested in the large $x$ behaviour. So I need to turn the sum into an asymptotic series. Using \begin{align} &\qquad \left(2\pi k + ix\right)^{s-1} - \left(2\pi k - ix\right)^{s-1} \\ &= \left(ix\right)^{s-1} \left(1 - \frac{i 2\pi k}{x}\right)^{s-1} - \left(-ix\right)^{s-1} \left(1 + \frac{i 2\pi k}{x}\right)^{s-1} \\ &= -ix^{s-1} \sum_{m=0}^\infty \begin{pmatrix} s-1 \\ m \end{pmatrix} \left\{ e^{i\frac{\pi s}{2}} \left(-\frac{i2\pi k}{x}\right)^m + e^{-i\frac{\pi s}{2}} \left(\frac{i2\pi k}{x}\right)^m \right\} \\ &= -2ix^{s-1} \sum_{m=0}^\infty \begin{pmatrix} s-1 \\ m \end{pmatrix} \cos\left(\frac{\pi(s-m)}{2}\right) \left(\frac{2\pi k}{x}\right)^m \\ &= -2i x^{s-1} \Bigg\{ \cos\left(\frac{\pi s}{2}\right) \sum_{m=0}^\infty \begin{pmatrix} s-1 \\ 2m \end{pmatrix} (-1)^m \left(\frac{2\pi k}{x}\right)^{2m} \\ &\quad + \sin\left(\frac{\pi s}{2}\right) \sum_{m=0}^\infty \begin{pmatrix} s-1 \\ 2m+1 \end{pmatrix} (-1)^m \left(\frac{2\pi k}{x}\right)^{2m+1} \Bigg\} \end{align} and changing $k$ and $m$ summation order to turn it into an asymptotic expansion, the function becomes \begin{align} f(x) &= 2x^{s-1} (2\pi)^{-s} \sum_{m=0}^\infty \begin{pmatrix} s-1 \\ m \end{pmatrix} \cos\left(\frac{\pi(s-m)}{2}\right) \left(\frac{2\pi}{x}\right)^m \zeta\left(s-m\right) \\ &= {2x^{s-1}}{(2\pi)^{-s}} \Bigg\{ \cos\left(\frac{\pi s}{2}\right) \sum_{m=0}^\infty \begin{pmatrix} s-1 \\ 2m \end{pmatrix} (-1)^m \left(\frac{2\pi}{x}\right)^{2m} \zeta\left(s-2m\right) \\ &\quad + \sin\left(\frac{\pi s}{2}\right) \sum_{m=0}^\infty \begin{pmatrix} s-1 \\ 2m+1 \end{pmatrix} (-1)^m \left(\frac{2\pi}{x}\right)^{2m+1} \zeta\left(s-2m-1\right) \Bigg\} \end{align} or to leading order \begin{align} f(x) &= {2x^{s-1}}{(2\pi)^{-s}} \left\{ \cos\left(\frac{\pi s}{2}\right) \, \zeta(s) + {\cal O}\left(\frac{1}{x}\right) \right\} \\ &= \frac{\zeta(1-s)}{\Gamma(s)} \, x^{s-1} + {\cal O}\left(x^{s-2}\right) \, . \end{align}