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

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Theorem

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

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

where:

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

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


Basis for the Induction

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

\(\displaystyle \) \(\) \(\displaystyle \sum_k \left\{ {0 + 1 \atop k + 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m}\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_k \delta_{1 \left({k + 1}\right)} \left[{k \atop m}\right] \left({-1}\right)^{k - m}\) Stirling Number of the Second 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
\(\displaystyle \) \(=\) \(\displaystyle \delta_{0 m} \left({-1}\right)^{- m}\) Unsigned Stirling Number of the First Kind of 0
\(\displaystyle \) \(=\) \(\displaystyle \delta_{0 m}\) multiplier irrelevant when $m \ne 0$
\(\displaystyle \) \(=\) \(\displaystyle \binom 0 m\) Zero Choose n

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:

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


from which it is to be shown that:

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


Induction Step

This is the induction step:

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


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


Therefore:

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

$\blacksquare$


Also see

Sources