Harmonic Number is not Integer

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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 1

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

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


Aiming for a contradiction, suppose otherwise:

$(\text P): \quad \exists m \in \N: H_m \in \Z$

By the definition of the harmonic numbers:

$H_m = 1 + \dfrac 1 2 + \dfrac 1 3 + \cdots + \dfrac 1 m$


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

Then:

\(\ds H_m\) \(=\) \(\ds 1 + \frac 1 2 + \frac 1 3 + \cdots + \frac 1 m\)
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds H_m - \frac 1 {2^t}\) \(=\) \(\ds 1 + \frac 1 2 + \frac 1 3 + \cdots + \frac 1 {2^t - 1} + \frac 1 {2^t + 1} + \cdots + \frac 1 m\)
\(\ds \leadsto \ \ \) \(\ds 2^{t - 1} H_m - \frac 1 2\) \(=\) \(\ds 2^{t - 1} + \frac {2^{t - 1} } 2 + \frac {2^{t - 1} } 3 + \frac {2^{t - 1} } 4 + \frac {2^{t - 1} } 5 + \frac {2^{t - 1} } 6 + \cdots + \frac {2^{t - 1} } m\) multiplying by $2^{t-1}$
\(\text {(2)}: \quad\) \(\ds \) \(=\) \(\ds 2^{t - 1} + 2^{t - 2} + \frac {2^{t - 1} } 3 + 2^{t - 3} + \frac {2^{t - 1} } 5 + \frac {2^{t - 2} } 3 + \cdots + \frac {2^{t - 1} } m\) cancelling powers of $2$


Let $S$ be the set of denominators on the right hand side 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 right hand side of $(1)$:

$\set {1, 2, 3, \ldots, 2^t - 1, 2^t + 1, \ldots, m}$

contains no multiple of $2^t$.

Therefore there can be no multiple of $2$ in the denominators of the right hand side of $(2)$.

Let:

$\ell = \map \lcm S$

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

Because $2$ is not a divisor of any of the elements of $S$, it will not be a divisor of $\ell$.

Hence $\ell$ is likewise odd.


We have:

\(\ds 2^{t - 1} H_m - \frac 1 2\) \(=\) \(\ds 2^{t - 1} + 2^{t - 2} + \frac {2^{t - 1} } 3 + 2^{t - 3} + \frac {2^{t - 1} } 5 + \frac {2^{t - 2} } 3 + \ldots + \frac {2^{t - 1} } m\) from $(2)$
\(\ds \leadsto \ \ \) \(\ds \frac {2^t H_m - 1} 2\) \(=\) \(\ds \frac {2^{t - 1} \ell + 2^{t - 2} \ell + 2^{t - 1} \paren {\ell / 3} + 2^{t - 3} \ell + 2^{t - 1} \paren {\ell / 5} + \ldots + 2^{t - 1} \paren {\ell / m} } \ell\) multiplying top and bottom by $\ell$
\(\ds \leadsto \ \ \) \(\ds \ell \paren {2^t H_m - 1}\) \(=\) \(\ds 2 \paren {2^{t - 1} \ell + 2^{t - 2} \ell + 2^{t - 1} \paren {\ell / 3} + 2^{t - 3} \ell + 2^{t - 1} \paren {\ell / 5} + \ldots + 2^{t - 1} \paren {\ell / m} }\) multiplying both sides by $2 \ell$


But the left hand side of that last equation is odd, while its right hand side is even.

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

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

$\blacksquare$


Proof 2

Aiming for a contradiction, suppose:

$(\text P): \quad \exists m \in \N: H_m \in \Z$

By the definition of the harmonic numbers:

$1 + \dfrac 1 2 + \dfrac 1 3 + \cdots +\dfrac 1 m = H_m$

$m$ is either prime or composite.


If $m$ is prime, we have that:

\(\ds 1 + \frac 1 2 + \frac 1 3 + \dots + \frac 1 m\) \(=\) \(\ds H_m\)
\(\ds m! + \frac {m!} 2 + \frac {m!} 3 + \cdots + \paren {m - 1}!\) \(=\) \(\ds m!\cdot H_m\) multiplying by $m!$

If $k \le n$, then $k$ divides $n!$, so all terms are integers.

Every term on the left hand side is divisible by $m$ except for one, namely $\paren {m - 1}!$.

Hence, the left hand side is not divisible by $m$.

But the right hand side is divisible by $m$, so this is a contradiction.


If $m$ is composite, let $p$ be the largest prime which is less than $m$.

Then we have:

\(\ds 1 + \frac 1 2 + \frac 1 3 + \dots + \frac 1 p + \cdots + \frac 1 m\) \(=\) \(\ds H_m\)
\(\ds m! + \frac {m!} 2 + \frac {m!} 3 + \cdots \frac {m!} p + \cdots + \paren {m - 1}!\) \(=\) \(\ds m! \cdot H_m\) multiplying by $m!$

For $k < p$, we have that $m!$ is divisible by $p$ and $k$ is not.

So from Euclid's Lemma, every term $\dfrac {m!} k$ is an integer divisible by $p$.


Aiming for a contradiction, suppose there exists $k \in \Z$, with $p< k< m$ such that $\dfrac {m!} k$ is not divisible by $p$.

Because $m!$ is divisible by $p$ and $\dfrac{m!} k$ is not, it follows from Euclid's Lemma that $k$ is a multiple of $p$.

Since $k$ is a multiple of $p$ which is greater than $p$:

$2 p \le k$

Thus:

$p < 2 p \le k < m$

From Bertrand's Postulate, there exists a prime $q$ such that $p < q < 2 p$.

Hence:

$p < q <m $

This contradicts the fact that $p$ is the largest prime less than $m$.


Hence, the assumption that there exists $k \in \Z$ with $p < k < m$ such that $\dfrac {m!} k$ is not divisible by $p$ is false.

Therefore, every term of the left hand side, except perhaps $\dfrac{m!} p$, is a multiple of $p$.


From above:

$p < m < 2 p$

so $p$ is the only positive integer less than $m$ which is divisible by $p$.

Since $m!$ is the product of positive integer less than or equal to $m$, it follows that $m!$ is divisible by $p$ only once.

Therefore, $\dfrac {m!} p$ is not divisible by $p$.

Because every term of the left hand side is divisible by $p$ except for one, it follows that the left hand side is not divisible by $p$.

Because the right hand side is a multiple of $m!$, and hence a multiple of $p$, this is a contradiction.

So the assumption that there exists $m \in \N: H_m \in \Z$ is false.

$\blacksquare$


Sources

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