Harmonic Number as Unsigned Stirling Number of First Kind over Factorial

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:

Thus $P \left({0}\right)$ is seen to hold.

Basis for the Induction
$P \left({1}\right)$ is the case:

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:

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \Z_{\ge 0}: H_n = \dfrac {\left[{ {n + 1} \atop 2}\right]} {n!}$