TITLE: Stirling Number of the Second Kind Identity
QUESTION [9 upvotes]: I'm aware of the identity
\begin{align}
\sum_{i = 0}^{k} i! \binom{n+1}{i + 1} S(k,i) = H_{n,-k},
\end{align}
where $H_{n,-k}$ is a generalized Harmonic number defined by $H_{n,m} = \sum_{r = 1}^{n} r^{-m}$. I believe the following sum is related to the generalized Harmonic numbers as well,
\begin{align}
\sum_{i = 0}^{k} (-1)^{i} i! \binom{n-i}{k-i} S(j,i)
\end{align}
and should be a nice function of $n, k$ and $j$, where $0 \leq j, k \leq n$. Any hints are certainly welcome!

REPLY [3 votes]: By way  of enrichment  of this discussion  we give a  purely algebraic
proof that does not use Eulerian numbers.
Suppose we seek to prove that
$$\sum_{q=0}^n q! {m+1\choose q+1} {n\brace q} = H_{m,-n}$$
with $m\ge n.$
Recall  the bivariate generating  function $G(z,  u)$ of  the Stirling
numbers of the second kind which is
$$G(z, u) = \exp(u(\exp(z)-1)).$$
This yields for the sum that
$$\sum_{q=0}^n q! {m+1\choose q+1} n! [z^n] [u^q] \exp(u(\exp(z)-1))
\\= n! [z^n] \sum_{q=0}^n q! {m+1\choose q+1} \frac{(\exp(z)-1)^q}{q!}
= n! [z^n] \sum_{q=0}^n {m+1\choose q+1} (\exp(z)-1)^q.$$
An important observation at this point is that $\exp(z)-1$ starts at
$z$, so  we may  extend the summation  to $m$ without  affecting the
coefficient of $z^n,$ getting
$$n! [z^n] \sum_{q=0}^m {m+1\choose q+1} (\exp(z)-1)^q.$$
Rewrite this as
$$(m+1) \times n! [z^n] 
\sum_{q=0}^m {m\choose q} \frac{(\exp(z)-1)^q}{q+1}
\\= (m+1) \times n! [z^n] \frac{1}{\exp(z)-1}
\sum_{q=0}^m {m\choose q} \frac{(\exp(z)-1)^{q+1}}{q+1}.$$
Stop for a moment to perform a formal power series integration.
We have $$(1+x)^m = \sum_{q=0}^m {m\choose q} x^q$$
so that
$$\frac{1}{m+1}(1+x)^{m+1} - \frac{1}{m+1}
=\frac{1}{m+1}\left((1+x)^{m+1} - 1\right)
\\= \sum_{q=0}^m {m\choose q} \frac{x^{q+1}}{q+1}.$$
Apply this to the sum to obtain
$$(m+1) \times n! [z^n] \frac{1}{\exp(z)-1}
\frac{1}{m+1}\left(\exp(z(m+1)) - 1\right)$$
which is
$$n! [z^n] \frac{\exp(z(m+1)) - 1}{\exp(z)-1}
= n! [z^n] \sum_{k=0}^m \exp(zk)
\\= n! \sum_{k=0}^m \frac{k^n}{n!} = \sum_{k=0}^m k^n = H_{m, -n}.$$