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

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Theorem

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

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

where:

$\displaystyle \left[{n + 1 \atop k + 1}\right]$ etc. denotes an unsigned Stirling number of the first kind
$\dbinom k m$ denotes a binomial coefficient.


Proof

The proof proceeds by induction.


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

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


Basis for the Induction

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

\(\displaystyle \sum_k \left[{1 \atop k + 1}\right] \binom k m \left({-1}\right)^{- m}\) \(=\) \(\displaystyle \sum_k \delta_{1 \left({k + 1}\right)} \binom k m \left({-1}\right)^{- m}\) Unsigned Stirling Number of the First Kind of 1
\(\displaystyle \) \(=\) \(\displaystyle \sum_k \delta_{0 k} \binom k m \left({-1}\right)^{- m}\) Definition of Kronecker Delta
\(\displaystyle \) \(=\) \(\displaystyle \binom 0 m \left({-1}\right)^{- m}\) all terms vanish except for $k = 0$
\(\displaystyle \) \(=\) \(\displaystyle \delta_{0 m} \left({-1}\right)^{- m}\) Zero Choose n
\(\displaystyle \) \(=\) \(\displaystyle \left[{0 \atop m}\right] \left({-1}\right)^{- m}\) Unsigned Stirling Number of the First Kind of 0

But when $m \ne 0$ we have that:

$\displaystyle \left[{0 \atop m}\right] \left({-1}\right)^{- m} = 0 = \left[{0 \atop m}\right]$

and when $m = 0$ we have that:

$\displaystyle \left[{0 \atop m}\right] \left({-1}\right)^{- 0} = 1 = \left[{0 \atop m}\right]$

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:

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


from which it is to be shown that:

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


Induction Step

This is the induction step:

\(\displaystyle \) \(\) \(\displaystyle \sum_k \left[{r + 2 \atop k + 1}\right] \binom k m \left({-1}\right)^{k - m}\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_k \left({\left({r + 1}\right) \left[{r + 1 \atop k + 1}\right] + \left[{r + 1 \atop k}\right]}\right) \binom k m \left({-1}\right)^{k - m}\) Definition of Unsigned Stirling Numbers of the First Kind
\(\displaystyle \) \(=\) \(\displaystyle \left({r + 1}\right) \sum_k \left[{r + 1 \atop k + 1}\right] \binom k m \left({-1}\right)^{k - m} + \sum_k \left[{r + 1 \atop k}\right] \binom k m \left({-1}\right)^{k - m}\)
\(\displaystyle \) \(=\) \(\displaystyle \left({r + 1}\right) \left[{r \atop m}\right] + \sum_k \left[{r + 1 \atop k}\right] \binom k m \left({-1}\right)^{k - m}\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle \left({r + 1}\right) \left[{r \atop m}\right] + \sum_k \left[{r + 1 \atop k}\right] \left({\binom {k - 1} m + \binom {k - 1} {m - 1} }\right) \left({-1}\right)^{k - m}\) Pascal's Rule
\(\displaystyle \) \(=\) \(\displaystyle \left({r + 1}\right) \left[{r \atop m}\right] + \sum_k \left[{r + 1 \atop k}\right] \binom {k - 1} m \left({-1}\right)^{k - m} + \sum_k \left[{r + 1 \atop k}\right] \binom {k - 1} {m - 1} \left({-1}\right)^{k - m}\)
\(\displaystyle \) \(=\) \(\displaystyle \left({r + 1}\right) \left[{r \atop m}\right] + \sum_k \left[{r + 1 \atop k + 1}\right] \binom k m \left({-1}\right)^{k + 1 - m} + \sum_k \left[{r + 1 \atop k + 1}\right] \binom k {m - 1} \left({-1}\right)^{k + 1 - m}\) Translation of Index Variable of Summation
\(\displaystyle \) \(=\) \(\displaystyle \left({r + 1}\right) \left[{r \atop m}\right] - \sum_k \left[{r + 1 \atop k + 1}\right] \binom k m \left({-1}\right)^{k - m} + \sum_k \left[{r + 1 \atop k + 1}\right] \binom k {m - 1} \left({-1}\right)^{k - \left({m - 1}\right)}\)
\(\displaystyle \) \(=\) \(\displaystyle \left({r + 1}\right) \left[{r \atop m}\right] - \left[{r \atop m}\right] + \left[{r \atop m - 1}\right]\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle r \left[{r \atop m}\right] + \left[{r \atop m - 1}\right]\)
\(\displaystyle \) \(=\) \(\displaystyle \left[{r + 1 \atop m}\right]\) Definition of Unsigned Stirling Numbers of the First Kind


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


Therefore:

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

$\blacksquare$


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