One usually encounters the harmonic series
\[
\sum_{k=1}^{\infty} \frac{1}{k}
\]
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} + \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.