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
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: $(51)$