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 all $n \ge 2$.

Seeking a contradiction, assume otherwise:
 * $(P): \quad \exists m \in \N: H_m \in \Z$

By the definition of the harmonic numbers:


 * $\displaystyle H_m = 1 + \frac 1 2 + \frac 1 3 + \ldots + \frac 1 m$

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

Then:

Let $S$ be the set of denominators on the RHS of $(2)$.

Then no element of $S$ can have $2$ as a factor, as follows.

Consider an arbitrary summand:


 * $\dfrac {2^{t-1}}{2^j \times k}$

for some $k \in \Z$, where $j \ge 0$ is the highest power of $2$ that divides the denominator.

For any $2$ to remain after simplification, we would need $j > t - 1$.

Were this to be so, then $2^j\times k$ would have $2^t$ as a factor, and some denominator would be a multiple of $2^t$.

By Greatest Power of Two not Divisor, the set of denominators of the RHS of $(1)$:


 * $\left\{ {1, 2, 3, \ldots, 2^t - 1, 2^t + 1, \ldots, m}\right\}$

contains no multiple of $2^t$.

Therefore there can be no multiple of $2$ in the denominators of the RHS of $(2)$.

Let:


 * $\ell = \operatorname{lcm} \left({S}\right)$

be the least common multiple of the elements of $S$.

Because all the elements of $S$ are odd, $\ell$ is likewise odd.

We have:

But the LHS of that last equation is odd, while its RHS is even.

As this is a contradiction, it follows that our assumption $(P)$ that such an $m$ exists is false.

That is, there are no harmonic numbers apart from $0$ and $1$ which are integers.