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

$\ds \sum_k {k + 1 \brace m + 1} \binom n k \paren {-1}^{n - k} = {n \brace m}$

where:

$\ds {k + 1 \brace m + 1}$ and so on 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 $\map P n$ be the proposition:

$\ds \forall m \in \Z_{\ge 0}: \sum_k {k + 1 \brace m + 1} \binom n k \paren {-1}^{n - k} = {n \brace m}$


Basis for the Induction

$\map P 0$ is the case:

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

So $\map P 0$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

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


So this is the induction hypothesis:

$\ds \sum_k {k + 1 \brace m + 1} \binom r k \paren {-1}^{r - k} = {r \brace m}$


from which it is to be shown that:

$\ds \sum_k {k + 1 \brace m + 1} \binom {r + 1} k \paren {-1}^{r + 1 - k} = {r + 1 \brace m}$


Induction Step

This is the induction step:

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


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


Therefore:

$\ds \forall m, n \in \Z_{\ge 0}: \sum_k {k + 1 \brace m + 1} \binom n k \paren {-1}^{n - k} = {n \brace m}$

$\blacksquare$


Sources

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