Harmonic Number is not Integer

Theorem
Let $H_n$ be the $n$th harmonic number.

Then $H_n$ is not an integer for $n \ge 2$.

That is, the only harmonic numbers that are integers are $H_0$ and $H_1$.

Proof
As $H_0 = 0$ and $H_1 = 1$, they are integers.

The claim is that $H_n$ is not an integer for $n \ge 2$.

Seeking a contradiction, assume otherwise.

By the definition of the harmonic numbers:


 * $H_n = 1 + \dfrac 1 2 + \dfrac 1 3 + \ldots + \dfrac 1 n$

Consider the set of denominators:


 * $D = \left \{ {1, 2, 3, \ldots, n} \right \}$

Let $t$ be the greatest power of two in $D$.

Then $H_n$ can be expressed as:


 * $(1): \quad H_n = 1 + \dfrac 1 2 + \dfrac 1 3 + \ldots + \dfrac 1 t + \ldots + \dfrac 1 n$

Let:


 * $\ell := \dfrac { \operatorname{lcm} \left({D}\right)} t$

where $\operatorname{lcm}$ denotes the least common multiple.

Multiplying both sides of $(1)$ by $\ell$ gives:


 * $\ell H_n = \ell + \dfrac \ell 2 + \dfrac \ell 3 + \ldots + \dfrac \ell t + \ldots + \dfrac \ell n$

From the assumption that $H_n$ is an integer, it follows that so too must $\ell H_n$ be.

By the definition of Least Common Multiple, every summand on the RHS except for $\dfrac \ell t$ is an integer.

From Greatest Power of Two not Divisor, $\dfrac \ell t$ is not an integer.

It follows that the RHS is not an integer.

But the LHS $H_n$ has been assumed to be an integer.

From this contradiction it follows that $H_n$ is not an integer.