# 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

## 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:

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

$\blacksquare$