Summation over Lower Index of Unsigned Stirling Numbers of the First Kind
Jump to navigation
Jump to search
Theorem
Let $n \in \Z_{\ge 0}$ be a positive integer.
Then:
- $\ds \sum_k {n \brack k} = n!$
where:
- $\ds {n \brack k}$ denotes an unsigned Stirling number of the first kind
- $n!$ denotes the factorial of $n$.
Proof
The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P N$ be the proposition:
- $\ds \sum_k {n \brack k} = n!$
$\map P 0$ is the case:
\(\ds \sum_k {0 \brack k}\) | \(=\) | \(\ds \sum_k \delta_{0 k}\) | Unsigned Stirling Number of the First Kind of 0 | |||||||||||
\(\ds \) | \(=\) | \(\ds 1\) | all terms vanish but for $k = 0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 0!\) | Definition of Factorial |
Thus $\map P 0$ is seen to hold.
Basis for the Induction
$\map P 1$ is the case:
\(\ds \sum_k {1 \brack k}\) | \(=\) | \(\ds \sum_k \delta_{1 k}\) | Unsigned Stirling Number of the First Kind of 1 | |||||||||||
\(\ds \) | \(=\) | \(\ds 1\) | all terms vanish but for $k = 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 1!\) | Definition of Factorial |
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that, if $\map P m$ is true, where $m \ge 2$, then it logically follows that $\map P {m + 1}$ is true.
So this is the induction hypothesis:
- $\ds \sum_k {m \brack k} = m!$
from which it is to be shown that:
- $\ds \sum_k {m + 1 \brack k} = \paren {m + 1}!$
Induction Step
This is the induction step:
\(\ds \sum_k {m + 1 \brack k}\) | \(=\) | \(\ds \sum_k \paren {m {m \brack k} + {m \brack k - 1} }\) | Definition of Unsigned Stirling Numbers of the First Kind | |||||||||||
\(\ds \) | \(=\) | \(\ds m \sum_k {m \brack k} + \sum_k {m \brack k - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds m \sum_k {m \brack k} + \sum_k {m \brack k}\) | Translation of Index Variable of Summation | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {m + 1} \sum_k {m \brack k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {m + 1} m!\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {m + 1}!\) | Definition of Factorial |
So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \forall n \in \Z_{\ge 0}: \sum_k {n \brack k} = n!$
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: Exercise $39$