I was wondering if the following series converges:
$$\sum_{k=0}^\infty\frac{\Gamma(r+2)}{k!\Gamma(r+2-k)}B_kn^{-k}\tag1$$
And if so, does it converge to
$$\frac{r+1}{n^{r+1}}\sum_{k=1}^nk^r\tag2$$
Here $B_k$ is the $k^{th}$ Bernoulli number and $\Gamma$ is the Gamma function. We also have $r\in\mathbb R$ and $n\in\mathbb N$.
It is just my intuition from Faulhaber's formula, but I don't know how to go about working it out, mainly because of $B_k$ and it is rather difficult to get such an expression into WA.
I don't mind if we start with $(2)$ and work backwards to $(1)$, but I still don't think I'm up for this problem.
If $r$ isn't an integer, the series doesn't converge.
Aside from $B_1$, all odd Bernoulli numbers vanish. So we could rewrite the series as $$ 1+\frac{B_1 \Gamma (r+2)}{n \Gamma (r+1)}+\sum_{k=1}^{\infty} \frac{\Gamma(r+2)}{(2k)! \Gamma(r+2-2k)}B_{2k}n^{-2k} $$Using the Ratio Test, we are interested in the limit $$ \lim_{k\to \infty}\left|\frac{1}{(2k+2)! \Gamma(r-2k)}B_{2k+2}n^{-2k-2}\cdot \frac{(2k)! \Gamma(r+2-2k)}{B_{2k}}n^{-2k}\right| $$ $$ =n^{-2}\lim_{k\to \infty}\left|\frac{(2 k-r-1) (2 k-r)}{ (2 k+1)(2k+2)}\cdot \frac{B_{2k+2}}{B_{2k}}\right| $$Using our assumption $r\notin \mathbb{N}$, we can simplify the rational part. $$ =1\cdot n^{-2}\lim_{k\to \infty}\left|\frac{B_{2k+2}}{B_{2k}}\right| $$Using the value of $\zeta(2k)$ and Stirling's formula, one can deduce the asymptotic $$ |B_{2k}|\sim \sqrt{4\pi k}\left(\frac{k}{\pi e}\right)^{2k} $$This gives $$ \lim_{k\to\infty}\left|\frac{B_{2k+2}}{B_{2k}}\right| = \lim_{k\to\infty}\left|\frac{k^{-2 k-\frac{1}{2}} (k+1)^{2 k+\frac{5}{2}}}{e^2 \pi ^2}\right|=\infty $$