The Harmonic Series Is Never an Integer

Why it's beautiful: $H_n = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}$ grows without bound, so you might expect it to eventually hit an integer. It never does (for $n \geq 2$ ). The proof uses a clever "lonely prime" argument: there is always one prime that appears exactly once in the denominators, and that prime poisons the entire sum, preventing it from being an integer.

Problem

Let $H_n = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}$ .

Prove that $H_n$ is not an integer for any $n \geq 2$ .

Hint: Let $p$ be the largest prime $\leq n$ . By Bertrand's postulate, $p > n/2$ , so $2p > n$ — meaning $p$ appears only once among $\{1, 2, \ldots, n\}$ . Now multiply $H_n$ by $L = \text{lcm}(1, 2, \ldots, n)$ and think about divisibility by $p$ .

Harmonic Series Is Never an Integer — Answer

Setup: The "Lonely Prime" Idea

Bertrand's Postulate states: for every integer $n \geq 1$ , there exists a prime $p$ with $n < p \leq 2n$ .

Equivalently: the largest prime $p \leq n$ satisfies $p > n/2$ , so $2p > n$ .

This means $p$ is the only multiple of $p$ in $\{1, 2, \ldots, n\}$ — hence "lonely."

Proof

Let $n \geq 2$ , and let $p$ be the largest prime $\leq n$ .

Step 1: Multiply through by $L = \text{lcm}(1, 2, \ldots, n)$ .

$L \cdot H_n = \frac{L}{1} + \frac{L}{2} + \cdots + \frac{L}{n}$

Every term $L/k$ is an integer (since $k \mid L$ by definition of lcm). So $L \cdot H_n$ is an integer.

If $H_n$ were also an integer, then $p \mid L \cdot H_n$ (since $p \mid L$ ). We will show this is impossible.

Step 2: Determine the exact power of $p$ in $L$ .

Since $p > n/2$ , we have $p^2 > n$ , so $p^2$ does not divide any $k \leq n$ . Therefore:

$p^1 \,\Big\|\, L \quad \text{(i.e., } p \mid L \text{ but } p^2 \nmid L\text{)}$

Step 3: Analyze each term $L/k$ modulo $p$ .

If $k \neq p$ : since $p \nmid k$ (because $p$ is the only multiple of $p$ in $\{1,\ldots,n\}$ ), and $p^1 \| L$ , we get $p \mid L/k$ .
If $k = p$ : since $p^1 \| L$ , removing one factor of $p$ gives $p \nmid L/p$ .

So modulo $p$ :

$L \cdot H_n = \underbrace{\frac{L}{p}}_{\not\equiv\, 0} + \underbrace{\sum_{k \neq p} \frac{L}{k}}_{\equiv\, 0} \not\equiv 0 \pmod{p}$

Step 4: Conclude.

We have shown $p \nmid L \cdot H_n$ .

But if $H_n = m$ were an integer, then $L \cdot H_n = L \cdot m$ , and since $p \mid L$ , we'd have $p \mid L \cdot m$ — contradiction.

Therefore $H_n$ is not an integer. $\square$

Why each step is needed

| Step | Role | |------|------| | Bertrand's postulate | Guarantees $p$ appears exactly once in $\{1,\ldots,n\}$ | | $p^1 \| L$ | Ensures $L/p$ is not divisible by $p$ , but all other $L/k$ are | | Mod $p$ analysis | Shows $L \cdot H_n \not\equiv 0 \pmod{p}$ , killing integrality |

The lonely prime $p$ is the obstruction. Every other denominator cancels cleanly — but $p$ has no partner to cancel with, so it leaves a remainder that can never be an integer.