# Sum over k of Stirling Numbers of the Second Kind of k with m by n choose k

## Theorem

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

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

where:

$\ds {k \brace m}$ 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:

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

### Basis for the Induction

$\map P 0$ is the case:

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

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:

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

from which it is to be shown that:

$\ds \sum_k {k \brace m} \binom {r + 1} k = {r + 2 \brace m + 1}$

### Induction Step

This is the induction step:

 $\ds$  $\ds \sum_k {k \brace m} \binom {r + 1} k$ $\ds$ $=$ $\ds \sum_k {k \brace m} \paren {\binom r k + \binom r {k - 1} }$ Pascal's Rule $\ds$ $=$ $\ds \sum_k {k \brace m} \binom r k + \sum_k {k \brace m} \binom r {k - 1}$ $\ds$ $=$ $\ds {r + 1 \brace m + 1} + \sum_k {k \brace m} \binom r {k - 1}$ Induction Hypothesis $\ds$ $=$ $\ds {r + 1 \brace m + 1} + \sum_k \paren {m {k - 1 \brace m} + {k - 1 \brace m - 1} } \binom r {k - 1}$ Definition of Stirling Numbers of the Second Kind $\ds$ $=$ $\ds {r + 1 \brace m + 1} + m \sum_k {k - 1 \brace m} \binom r {k - 1} + \sum_k {k - 1 \brace m - 1} \binom r {k - 1}$ $\ds$ $=$ $\ds {r + 1 \brace m + 1} + m \sum_k {k \brace m} \binom r k + \sum_k {k \brace m - 1} \binom r k$ Translation of Index Variable of Summation $\ds$ $=$ $\ds {r + 1 \brace m + 1} + m {r + 1 \brace m + 1} + {r + 1 \brace m}$ Induction Hypothesis $\ds$ $=$ $\ds \paren {m + 1} {r + 1 \brace m + 1} + {r + 1 \brace m}$ $\ds$ $=$ $\ds {r + 2 \brace m + 1}$ 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:

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

$\blacksquare$