Sum over k of Unsigned Stirling Number of the First Kind of n+1 with k+1 by Stirling Number of the Second Kind of k with m by -1^k-m

Theorem

Let $m, n \in \Z_{\ge 0}$.

$\displaystyle \sum_k \left[{n + 1 \atop k + 1}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{k - m} = \left[{n \ge m}\right] \dfrac {n!} {m!}$

where:

$\left[{n \ge m}\right]$ is Iverson's convention
$\displaystyle \left[{n + 1 \atop k + 1}\right]$ denotes an unsigned Stirling number of the first kind
$\displaystyle \left\{ {k \atop m}\right\}$ denotes a Stirling number of the second kind.

Proof

The proof proceeds by induction on $n$.

For all $n \in \Z_{\ge 0}$, let $P \left({m}\right)$ be the proposition:

$\displaystyle \sum_k \left[{n + 1 \atop k + 1}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{k - m} = \left[{n \ge m}\right] \dfrac {n!} {m!}$

Basis for the Induction

$P \left({0}\right)$ is the case:

 $\displaystyle$  $\displaystyle \sum_k \left[{0 + 1 \atop k + 1}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{k - m}$ $\displaystyle$ $=$ $\displaystyle \sum_k \delta_{1 \left({k + 1}\right)} \left\{ {k \atop m}\right\} \left({-1}\right)^{k - m}$ Unsigned Stirling Number of the First Kind of 1 $\displaystyle$ $=$ $\displaystyle \sum_k \delta_{0 k} \left\{ {k \atop m}\right\} \left({-1}\right)^{k - m}$ Definition of Kronecker Delta $\displaystyle$ $=$ $\displaystyle \left\{ {0 \atop m}\right\} \left({-1}\right)^{-m}$ All terms but where $k = 0$ vanish $\displaystyle$ $=$ $\displaystyle \delta_{0 m} \left({-1}\right)^{-m}$ Stirling Number of the Second Kind of 0 $\displaystyle$ $=$ $\displaystyle \delta_{0 m}$ multiplier irrelevant when $m \ne 0$ $\displaystyle$ $=$ $\displaystyle \left[{0 \ge 0}\right] \dfrac {0!} {0!}$ Factorial of Zero

So $P \left({0}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis

Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 1$, then it logically follows that $P \left({r + 1}\right)$ is true.

So this is the induction hypothesis:

$\displaystyle \sum_k \left[{r + 1 \atop k + 1}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{k - m} = \left[{r \ge m}\right] \dfrac {r!} {m!}$

from which it is to be shown that:

$\displaystyle \sum_k \left[{r + 2 \atop k + 1}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{k - m} = \left[{r + 1 \ge m}\right] \dfrac {\left({r + 1}\right)!} {m!}$

Induction Step

This is the induction step:

 $\displaystyle$  $\displaystyle \sum_k \left[{r + 2 \atop k + 1}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{k - m}$ $\displaystyle$ $=$ $\displaystyle \sum_k \left({\left({r + 1}\right) \left[{r + 1 \atop k + 1}\right] + \left[{r + 1 \atop k}\right] }\right) \left\{ {k \atop m}\right\} \left({-1}\right)^{k - m}$ Definition of Unsigned Stirling Numbers of the First Kind $\displaystyle$ $=$ $\displaystyle \left({r + 1}\right) \sum_k \left[{r + 1 \atop k + 1}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{k - m} + \sum_k \left[{r + 1 \atop k}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{k - m}$ $\displaystyle$ $=$ $\displaystyle \left[{r \ge m}\right] \dfrac {\left({r + 1}\right) r!} {m!} + \sum_k \left[{r + 1 \atop k}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{k - m}$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \left[{r \ge m}\right] \dfrac {\left({r + 1}\right)!} {m!} + \left({-1}\right)^{r + 1 - m} \sum_k \left[{r + 1 \atop k}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{r + 1 - k}$ $\displaystyle$ $=$ $\displaystyle \left[{r \ge m}\right] \dfrac {\left({r + 1}\right)!} {m!} + \left({-1}\right)^{r + 1 - m} \delta_{m \left({r + 1}\right)}$ First Inversion Formula for Stirling Numbers

So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_k \left[{n + 1 \atop k + 1}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{k - m} = \left[{n \ge m}\right] \dfrac {n!} {m!}$

$\blacksquare$