Sum over k of Unsigned Stirling Numbers of the First Kind of n with k by k choose m

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Theorem

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

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

where:

$\displaystyle \left[{n \atop k}\right]$ 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:

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


Basis for the Induction

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

\(\displaystyle \sum_k \left[{0 \atop k}\right] \binom k m\) \(=\) \(\displaystyle \sum_k \delta_{0 k} \binom k m\) Unsigned Stirling Number of the First Kind of 0
\(\displaystyle \) \(=\) \(\displaystyle \binom 0 m\) all terms vanish except for $k = 0$
\(\displaystyle \) \(=\) \(\displaystyle \delta_{0 m}\) Zero Choose n
\(\displaystyle \) \(=\) \(\displaystyle \delta_{1 \left({m + 1}\right)}\) Definition of Kronecker Delta
\(\displaystyle \) \(=\) \(\displaystyle \left[{1 \atop m + 1}\right]\) Unsigned Stirling Number of the First Kind of 1
\(\displaystyle \) \(=\) \(\displaystyle \left[{0 + 1 \atop m + 1}\right]\)

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:

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


from which it is to be shown that:

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


Induction Step

This is the induction step:

\(\displaystyle \) \(\) \(\displaystyle \sum_k \left[{r + 1 \atop k}\right] \binom k m\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_k \left({r \left[{r \atop k}\right] + \left[{r \atop k - 1}\right]}\right) \binom k m\) Definition of Unsigned Stirling Numbers of the First Kind
\(\displaystyle \) \(=\) \(\displaystyle r \sum_k \left[{r \atop k}\right] \binom k m + \sum_k \left[{r \atop k - 1}\right] \binom k m\)
\(\displaystyle \) \(=\) \(\displaystyle r \left[{r + 1 \atop m + 1}\right] + \sum_k \left[{r \atop k - 1}\right] \binom k m\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle r \left[{r + 1 \atop m + 1}\right] + \sum_k \left[{r \atop k - 1}\right] \left({\binom {k - 1} m + \binom {k - 1} {m - 1} }\right)\) Pascal's Rule
\(\displaystyle \) \(=\) \(\displaystyle r \left[{r + 1 \atop m + 1}\right] + \sum_k \left[{r \atop k - 1}\right] \binom {k - 1} m + \sum_k \left[{r \atop k - 1}\right] \binom {k - 1} {m - 1}\)
\(\displaystyle \) \(=\) \(\displaystyle r \left[{r + 1 \atop m + 1}\right] + \sum_k \left[{r \atop k}\right] \binom k m + \sum_k \left[{r \atop k}\right] \binom k {m - 1}\) Translation of Index Variable of Summation
\(\displaystyle \) \(=\) \(\displaystyle r \left[{r + 1 \atop m + 1}\right] + \left[{r + 1 \atop m + 1}\right] + \left[{r + 1 \atop m}\right]\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle \left({r + 1}\right) \left[{r + 1 \atop m + 1}\right] + \left[{r + 1 \atop m}\right]\)
\(\displaystyle \) \(=\) \(\displaystyle \left[{r + 2 \atop m + 1}\right]\) 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:

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

$\blacksquare$


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