# 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:

 $\displaystyle H_0$ $=$ $\displaystyle 0$ $\displaystyle$ $=$ $\displaystyle \dfrac {\left[{1 \atop 2}\right]} {0!}$ 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:

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

$\blacksquare$