Sum over k of Unsigned Stirling Numbers of the First Kind of n+1 with k+1 by k choose m by -1^k-m

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Theorem

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

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

where:

$\ds {n + 1 \brack k + 1}$ etc. 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:

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


Basis for the Induction

$\map P 0$ is the case:

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

But when $m \ne 0$ we have that:

$\ds {0 \brack m} \paren {-1}^{- m} = 0 = {0 \brack m}$

and when $m = 0$ we have that:

$\ds {0 \brack m} \paren {-1}^{- 0} = 1 = {0 \brack m}$

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:

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


from which it is to be shown that:

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


Induction Step

This is the induction step:

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

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

$\blacksquare$


Sources