A Simple Proof that the Harmonic Series Diverges

One usually encounters the harmonic series

\[\begin{equation*} \sum_{k=1}^{\infty} \frac{1}{k} \end{equation*}\]

as an example of a series that diverges for non-obvious reasons. With \(H_n\) defined to be the partial sum \(H_n:=\sum_{k=1}^n \frac{1}{k}\), a typical way to prove divergence is to observe that

\[\begin{align*} H_{2n} &= H_n + \frac{1}{n+1} + \dots + \frac{1}{2n} \\ & \geq H_n + \underbrace{\frac{1}{2n} + \dots + \frac{1}{2n}}_{n \text{ terms}} \\ & = H_n + \frac{1}{2}. \end{align*}\]

In particular, \(H_{2^n} \geq 1 + \frac{n}{2}\), so the sequence of partial sums of the harmonic series has an unbounded subsequence, whence the series diverges.

The other day the author stumbled upon a simple proof of divergence. Almost certainly this is not new, due to the simplicity of the argument. Suppose for the sake of a contradiction that the harmonic series converges to a number \(h\). Then

\[\begin{align*} h &= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}\dots \\ & \geq 1 + \frac{1}{2} + \left( \frac{1}{4} + \frac{1}{4} \right) + \left( \frac{1}{6} + \frac{1}{6} \right) + \dots \\ &= \frac{1}{2} + h, \end{align*}\]

an obvious contradiction.