# 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

## 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:

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

$\blacksquare$