Harmonic Number as Unsigned Stirling Number of First Kind over Factorial
Jump to navigation
Jump to search
Contents
Theorem
- $H_n = \dfrac {\left[{ {n + 1} \atop 2}\right]} {n!}$
where:
- $H_n$ denotes the $n$th harmonic number
- $n!$ denotes the $n$th factorial
- $\displaystyle \left[{ {n + 1} \atop 2}\right]$ denotes an unsigned Stirling number of the first kind.
Proof
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
- $H_n = \dfrac {\left[{ {n + 1} \atop 2}\right]} {n!}$
$P \left({0}\right)$ is the case:
\(\displaystyle H_0\) | \(=\) | \(\displaystyle 0\) | |||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {\left[{1 \atop 2}\right]} {0!}\) | Unsigned Stirling Number of the First Kind of Number with Greater |
Thus $P \left({0}\right)$ is seen to hold.
Basis for the Induction
$P \left({1}\right)$ is the case:
\(\displaystyle H_1\) | \(=\) | \(\displaystyle 1\) | |||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {\left[{2 \atop 2}\right]} {1!}\) | Unsigned Stirling Number of the First Kind of Number with Self |
Thus $P \left({1}\right)$ is seen to hold.
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true.
So this is the induction hypothesis:
- $H_k = \dfrac {\left[{ {k + 1} \atop 2}\right]} {k!}$
from which it is to be shown that:
- $H_{k + 1} = \dfrac {\left[{ {k + 2} \atop 2}\right]} {\left({k + 1}\right)!}$
Induction Step
This is the induction step:
\(\displaystyle \dfrac {\left[{ {k + 2} \atop 2}\right]} {\left({k + 1}\right)!}\) | \(=\) | \(\displaystyle \dfrac {\left({k + 1}\right) \left[{ {k + 1} \atop 2}\right] + \left[{ {k + 1} \atop 1}\right]} {\left({k + 1}\right)!}\) | Definition of Unsigned Stirling Numbers of the First Kind | ||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {\left({k + 1}\right) \left[{ {k + 1} \atop 2}\right] + k!} {\left({k + 1}\right)!}\) | Unsigned Stirling Number of the First Kind of n+1 with 1 | ||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {\left[{ {k + 1} \atop 2}\right]} {k!} + \dfrac 1 {k + 1}\) | simplifying | ||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle H_k + \dfrac 1 {k + 1}\) | Induction Hypothesis | ||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle H_{k + 1}\) | Definition of Harmonic Number |
So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall n \in \Z_{\ge 0}: H_n = \dfrac {\left[{ {n + 1} \atop 2}\right]} {n!}$
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.7$: Harmonic Numbers: Exercise $6$