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

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

where:

$\ds {n \brack k}$ 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 $\map P n$ be the proposition:

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


Basis for the Induction

$\map P 0$ is the case:

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

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:

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


from which it is to be shown that:

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


Induction Step

This is the induction step:

\(\ds \) \(\) \(\ds \sum_k {r + 1 \brack k} \binom k m\)
\(\ds \) \(=\) \(\ds \sum_k \paren {r {r \brack k} + {r \brack k - 1} } \binom k m\) Definition of Unsigned Stirling Numbers of the First Kind
\(\ds \) \(=\) \(\ds r \sum_k {r \brack k} \binom k m + \sum_k {r \brack k - 1} \binom k m\)
\(\ds \) \(=\) \(\ds r {r + 1 \brack m + 1} + \sum_k {r \brack k - 1} \binom k m\) Induction Hypothesis
\(\ds \) \(=\) \(\ds r {r + 1 \brack m + 1} + \sum_k {r \brack k - 1} \paren {\binom {k - 1} m + \binom {k - 1} {m - 1} }\) Pascal's Rule
\(\ds \) \(=\) \(\ds r {r + 1 \brack m + 1} + \sum_k {r \brack k - 1} \binom {k - 1} m + \sum_k {r \brack k - 1} \binom {k - 1} {m - 1}\)
\(\ds \) \(=\) \(\ds r {r + 1 \brack m + 1} + \sum_k {r \brack k} \binom k m + \sum_k {r \brack k} \binom k {m - 1}\) Translation of Index Variable of Summation
\(\ds \) \(=\) \(\ds r {r + 1 \brack m + 1} + {r + 1 \brack m + 1} + {r + 1 \brack m}\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \paren {r + 1} {r + 1 \brack m + 1} + {r + 1 \brack m}\)
\(\ds \) \(=\) \(\ds {r + 2 \brack m + 1}\) Definition of Unsigned Stirling Numbers of the First Kind


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


Therefore:

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

$\blacksquare$


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