Sum over k to n of Unsigned Stirling Number of the First Kind of k with m by n factorial over k factorial

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Theorem

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

$\ds \sum_{k \mathop \le n} {k \brack m} \frac {n!} {k!} = {n + 1 \brack m + 1}$

where:

$\ds {k \brack m}$ denotes an unsigned Stirling number of the first kind
$ n!$ denotes a factorial.


Proof

The proof proceeds by induction on $n$.


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

$\ds \forall n \in \Z_{\ge 0}: \sum_{k \mathop \le n} {k \brack m} \frac {n!} {k!} = {n + 1 \brack m + 1}$


Basis for the Induction

$\map P 0$ is the case:

\(\ds \) \(\) \(\ds \sum_{k \mathop \le 0} {k \brack m} \frac {0!} {k!}\)
\(\ds \) \(=\) \(\ds {0 \brack m} \frac 1 {0!}\) Definition of Unsigned Stirling Numbers of the First Kind
\(\ds \) \(=\) \(\ds \delta_{0 m}\) Unsigned Stirling Number of the First Kind of 0
\(\ds \) \(=\) \(\ds \delta_{1 \paren {m + 1} }\)
\(\ds \) \(=\) \(\ds {1 \brack m + 1}\) Unsigned Stirling Number of the First Kind of 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:

$\ds \sum_{k \mathop \le r} {k \brack m} \frac {r!} {k!} = {r + 1 \brack m + 1}$


from which it is to be shown that:

$\ds \sum_{k \mathop \le r + 1} {k \brack m} \frac {\paren {r + 1}!} {k!} = {r + 2 \brack m + 1}$


Induction Step

This is the induction step:

\(\ds \) \(\) \(\ds \sum_{k \mathop \le r + 1} {k \brack m} \frac {\paren {r + 1}!} {k!}\)
\(\ds \) \(=\) \(\ds {r + 1 \brack m} \frac {\paren {r + 1}!} {\paren {r + 1}!} + \paren {r + 1} \paren {\sum_{k \mathop \le r} {k \brack m} \frac {r!} {k!} }\)
\(\ds \) \(=\) \(\ds {r + 1 \brack m} + \paren {r + 1} {r + 1 \brack m + 1}\)
\(\ds \) \(=\) \(\ds {r + 2 \brack m + 1}\) 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:

$\ds \forall m, n \in \Z_{\ge 0}: \sum_{k \mathop \le n} {k \brack m} \frac {n!} {k!} = {n + 1 \brack m + 1}$

$\blacksquare$


Sources

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