# Riemann Zeta Function, Stirling's Numbers, and Infinite Series

A while back I was able to prove the following identity, $$\sum_{k=1}^{\infty}\frac{\Gamma(k+r)}{\Gamma(k)(k+r)^s}=\sum_{k=0}^{r}s(r+1,r+1-k)\zeta(s-r+k)$$ where $s(k,n)$ are the Sterling numbers of the first kind. To my satisfaction (and dismay) I was able to type this infinite series into WolframAlpha and it gave me my exact same finite summation to the point, meaning one I was correct but two it had already been done. After some googling I was seeing quite a few papers that tied finite sums involving Sterlings numbers and the Riemann Zeta function to infinite sums involving the Pochhammer symbol, $(x)_k$, $$(x)_k=\frac{\Gamma(x+k)}{\Gamma(x)}$$ I was wondering if anyone knew whether these results are well known and/or what central theorem/idea they stem from.

• Are you using the gamma function because you expect $r\in\mathbb{R}$? If so, I will point out that defining the Stirling numbers for fractional values is a pain, and the RHS summation has issues with its upper bound as well. If not, why not use the factorial? – Simply Beautiful Art Jul 2 '16 at 12:38

Suppose we seek to simplify the sum

$$\sum_{k=1}^\infty \frac{\Gamma(k+r)}{\Gamma(k) (k+r)^s}.$$

with $r$ a positive integer. This is

$$\sum_{k=1}^\infty \frac{(k+r-1)!}{(k-1)! (k+r)^s} = \sum_{k=1}^\infty \frac{1}{(k+r)^s} (k+r-1)^{\underline{r}} \\ = \sum_{k=1}^\infty \frac{1}{(k+r)^{s+1}} (k+r)^{\underline{r+1}}.$$

Now recall that

$$x^{\underline{r}} = \sum_{q=0}^r (-1)^{r-q} \left[r\atop q\right] x^q.$$

We thus obtain

$$\sum_{k=1}^\infty \frac{1}{(k+r)^{s+1}} \sum_{q=0}^{r+1} (-1)^{r+1-q} \left[r+1\atop q\right] (k+r)^q \\ = \sum_{q=0}^{r+1} (-1)^{r+1-q} \left[r+1\atop q\right] \left(\zeta(s+1-q) - \sum_{p=1}^r \frac{1}{p^{s+1-q}}\right).$$

Observe however that

$$\sum_{q=0}^{r+1} (-1)^{r+1-q} \left[r+1\atop q\right] \sum_{p=1}^r \frac{1}{p^{s+1-q}} \\ = \sum_{p=1}^r \frac{1}{p^{s+1}} \sum_{q=0}^{r+1} (-1)^{r+1-q} \left[r+1\atop q\right] p^q \\ = \sum_{p=1}^r \frac{1}{p^{s+1}} p^{\underline{r+1}}.$$

However with $1\le p\le r$ the term $p^{\underline{r+1}}$ includes a zero factor and hence the latter sum vanishes, leaving just

$$\sum_{q=0}^{r+1} (-1)^{r+1-q} \left[r+1\atop q\right] \zeta(s+1-q).$$

To ensure convergence we must have $s-r\gt 1.$

Wikipedia lists the Stirling number identity.

• Is that the rising factorial you have going over there? ($r$ in second line) – Simply Beautiful Art Jul 2 '16 at 12:36