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

## Theorem

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

$\displaystyle \sum_k {n + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {n \ge m} \dfrac {n!} {m!}$

where:

$\sqbrk {n \ge m}$ is Iverson's convention
$\displaystyle {n + 1 \brack k + 1}$ denotes an unsigned Stirling number of the first kind
$\displaystyle {k \brace m}$ denotes a Stirling number of the second kind.

## Proof

The proof proceeds by induction on $n$.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

$\displaystyle \sum_k {n + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {n \ge m} \dfrac {n!} {m!}$

### Basis for the Induction

$\map P 0$ is the case:

 $\displaystyle$  $\displaystyle \sum_k {n + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m}$ $\displaystyle$ $=$ $\displaystyle \sum_k \delta_{1 \paren {k + 1} } {k \brace m} \paren {-1}^{k - m}$ Unsigned Stirling Number of the First Kind of 1 $\displaystyle$ $=$ $\displaystyle \sum_k \delta_{0 k} {k \brace m} \paren {-1}^{k - m}$ Definition of Kronecker Delta $\displaystyle$ $=$ $\displaystyle {0 \brace m} \paren {-1}^{- m}$ All terms but where $k = 0$ vanish $\displaystyle$ $=$ $\displaystyle \delta_{0 m} \paren {-1}^{- m}$ Stirling Number of the Second Kind of 0 $\displaystyle$ $=$ $\displaystyle \delta_{0 m}$ multiplier irrelevant when $m \ne 0$ $\displaystyle$ $=$ $\displaystyle \sqbrk {0 \ge m} \dfrac {0!} {m!}$ Factorial of Zero

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:

$\displaystyle \sum_k \sqbrk {r + 1 \atop k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {r \ge m} \dfrac {r!} {m!}$

from which it is to be shown that:

$\displaystyle \sum_k \sqbrk {r + 2 \atop k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {r + 1 \ge m} \dfrac {\paren {r + 1}!} {m!}$

### Induction Step

This is the induction step:

 $\displaystyle$  $\displaystyle \sum_k \sqbrk {r + 2 \atop k + 1} {k \brace m} \paren {-1}^{k - m}$ $\displaystyle$ $=$ $\displaystyle \sum_k \paren {\paren {r + 1} {r + 1 \brack k + 1} + {r + 1 \brack k} } {k \brace m} \paren {-1}^{k - m}$ Definition of Unsigned Stirling Numbers of the First Kind $\displaystyle$ $=$ $\displaystyle \paren {r + 1} \sum_k {r + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m} + \sum_k {r + 1 \brack k} {k \brace m} \paren {-1}^{k - m}$ $\displaystyle$ $=$ $\displaystyle \sqbrk {r \ge m} \dfrac {\paren {r + 1} r!} {m!} + \sum_k {r + 1 \brack k} {k \brace m} \paren {-1}^{k - m}$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \sqbrk {r \ge m} \dfrac {\paren {r + 1}!} {m!} + \paren {-1}^{r + 1 - m} \sum_k {r + 1 \brack k} {k \brace m} \paren {-1}^{r + 1 - k}$ $\displaystyle$ $=$ $\displaystyle \sqbrk {r \ge m} \dfrac {\paren {r + 1}!} {m!} + \paren {-1}^{r + 1 - m} \delta_{m \paren {r + 1} }$ First Inversion Formula for Stirling Numbers $\displaystyle$ $=$ $\displaystyle \cases {\dfrac {\paren {r + 1}!} {m!} & : r + 1 > m \cr 1 & : r + 1 = m \cr 0 & : r + 1 < m}$ $\displaystyle$ $=$ $\displaystyle \sqbrk {r + 1 \ge m} \dfrac {\paren {r + 1}!} {m!}$

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

Therefore:

$\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_k {n + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {n \ge m} \dfrac {n!} {m!}$

$\blacksquare$