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:


 * $\displaystyle H_n = 1 + \frac 1 2 + \frac 1 3 + \ldots + \frac 1 n$

Let $2^t$ denote the highest power of two in the denominators of the summands:


 * $\displaystyle (1): H_n - \frac 1 {2^t} = 1 + \frac 1 2 + \frac 1 3 + \ldots + \frac 1 {2^t - 1} + \frac 1 {2^t + 1} + \ldots + \frac 1 n$

Multiply both sides of $(1)$ by $2^{t-1}$:


 * $\displaystyle (2): \quad 2^{t-1} H_n - \frac 1 2 = 2^{t-1} + 2^{t-2} + \frac {2^{t-1}} 3 + 2^{t-3} + \frac {2^{t-1}} 5 + \ldots + \frac {2^{t-1}} n$

Let all summands be written in canonical form.

Then no denominator on the RHS can have $2$ as a factor.

This is because, were this to be so, the denominator would have to have as a factor $2 \times 2^{t-1} = 2^t$.

That would contradict Greatest Power of Two not Divisor.

Let:


 * $\ell = \operatorname{lcm}\left\{{1, 3, 5, 7, \ldots, n}\right\}$

be the least common multiple of the denominators on the RHS.

Coalescing both sides of $(2)$:


 * $\displaystyle \frac {2^t H_n - 2}{2^t} = \frac {\ell 2^{t-1} + \ell 2^{t-2} + \ldots + \ell}{1 \times 3 \times 5 \times \ldots \times n}$

But this is a contradiction, as the denominator on the right does not have two as a factor.