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

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Theorem

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

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

where:

$\displaystyle \left\{ {k + 1 \atop m + 1}\right\}$ etc. denotes a Stirling number of the second kind
$\dbinom n k$ 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:

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


Basis for the Induction

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

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

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:

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


from which it is to be shown that:

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


Induction Step

This is the induction step:

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


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


Therefore:

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

$\blacksquare$


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