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

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Theorem

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

$\displaystyle \sum_k {n + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {n \ge m} \dfrac {n!} {m!}$

where:

$\sqbrk {n \ge m}$ is Iverson's convention
$\displaystyle {n + 1 \brack k + 1}$ denotes an unsigned Stirling number of the first kind
$\displaystyle {k \brace m}$ denotes a Stirling number of the second kind.


Proof

The proof proceeds by induction on $n$.


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

$\displaystyle \sum_k {n + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {n \ge m} \dfrac {n!} {m!}$


Basis for the Induction

$\map P 0$ is the case:

\(\displaystyle \) \(\) \(\displaystyle \sum_k {n + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m}\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_k \delta_{1 \paren {k + 1} } {k \brace m} \paren {-1}^{k - m}\) Unsigned Stirling Number of the First Kind of 1
\(\displaystyle \) \(=\) \(\displaystyle \sum_k \delta_{0 k} {k \brace m} \paren {-1}^{k - m}\) Definition of Kronecker Delta
\(\displaystyle \) \(=\) \(\displaystyle {0 \brace m} \paren {-1}^{- m}\) All terms but where $k = 0$ vanish
\(\displaystyle \) \(=\) \(\displaystyle \delta_{0 m} \paren {-1}^{- m}\) Stirling Number of the Second Kind of 0
\(\displaystyle \) \(=\) \(\displaystyle \delta_{0 m}\) multiplier irrelevant when $m \ne 0$
\(\displaystyle \) \(=\) \(\displaystyle \sqbrk {0 \ge m} \dfrac {0!} {m!}\) Factorial of Zero

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:

$\displaystyle \sum_k \sqbrk {r + 1 \atop k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {r \ge m} \dfrac {r!} {m!}$


from which it is to be shown that:

$\displaystyle \sum_k \sqbrk {r + 2 \atop k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {r + 1 \ge m} \dfrac {\paren {r + 1}!} {m!}$


Induction Step

This is the induction step:

\(\displaystyle \) \(\) \(\displaystyle \sum_k \sqbrk {r + 2 \atop k + 1} {k \brace m} \paren {-1}^{k - m}\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_k \paren {\paren {r + 1} {r + 1 \brack k + 1} + {r + 1 \brack k} } {k \brace m} \paren {-1}^{k - m}\) Definition of Unsigned Stirling Numbers of the First Kind
\(\displaystyle \) \(=\) \(\displaystyle \paren {r + 1} \sum_k {r + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m} + \sum_k {r + 1 \brack k} {k \brace m} \paren {-1}^{k - m}\)
\(\displaystyle \) \(=\) \(\displaystyle \sqbrk {r \ge m} \dfrac {\paren {r + 1} r!} {m!} + \sum_k {r + 1 \brack k} {k \brace m} \paren {-1}^{k - m}\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle \sqbrk {r \ge m} \dfrac {\paren {r + 1}!} {m!} + \paren {-1}^{r + 1 - m} \sum_k {r + 1 \brack k} {k \brace m} \paren {-1}^{r + 1 - k}\)
\(\displaystyle \) \(=\) \(\displaystyle \sqbrk {r \ge m} \dfrac {\paren {r + 1}!} {m!} + \paren {-1}^{r + 1 - m} \delta_{m \paren {r + 1} }\) First Inversion Formula for Stirling Numbers
\(\displaystyle \) \(=\) \(\displaystyle \cases {\dfrac {\paren {r + 1}!} {m!} & : $r + 1 > m$ \cr 1 & : $r + 1 = m$ \cr 0 & : $r + 1 < m$}\)
\(\displaystyle \) \(=\) \(\displaystyle \sqbrk {r + 1 \ge m} \dfrac {\paren {r + 1}!} {m!}\)


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


Therefore:

$\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_k {n + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {n \ge m} \dfrac {n!} {m!}$

$\blacksquare$


Also see


Sources