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

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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:

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

$\blacksquare$


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