What are different ways to prove that $\sum_{n=1}^{\infty}\frac 1n$ is divergent?

The slickest proof I have seen is the following.

We will argue by contradiction, assume that $\sum_{n=1}^{\infty}\dfrac{1}{n}=C <\infty.$ Then $\dfrac{C}{2}=\sum_{n=1}^{\infty}\dfrac{1}{2n}$, however

\begin{align} C=\sum_{n=1}^{\infty}\dfrac{1}{n} &=\sum_{n=1}^{\infty}\dfrac{1}{2n}+\sum_{n=0}^{\infty}\dfrac{1}{2n+1}\\ &>\sum_{n=1}^{\infty}\dfrac{1}{2n}+\sum_{n=0}^{\infty}\dfrac{1}{2n+2}\\ &=2\sum_{n=1}^{\infty}\dfrac{1}{2n} \end{align} Therefore, $$\dfrac{C}{2}>\sum_{n=1}^{\infty}\dfrac{1}{2n}$$ a contradiction.

Here is a proof that uses the fact that if a sequence $a_n$ converges, then $a_{2n}$ converges to the same value.

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Let $S_n$ be the sequence given by $S_n=\sum_{k=n+1}^{2n}\frac{1}{k}$. Then, note that

$$S_{n+1}-S_n=\frac{1}{2n+2}+\frac1{2n+1}>0$$

Therefore, the sequence $S_n$ is increasing. Furthermore, $S_n$ satisfies the estimates

$$\frac12=S_n<S_n=\sum_{k=n+1}^{2n}\frac{1}{k}\le \sum_{k=n+1}^{2n}\frac{1}{n+1}=\frac{n}{n+1}<1$$

Therefore, inasmuch as $S_n$ is increasing and bounded above, $S_n$ converges.

Noting that $S_n=\sum_{k=1}^{2n}\frac1k -\sum_{k=1}^n\frac1k$ converges to a number between $1/2$ and $1$ (i.e., not $0$), we conclude that $\sum_{k=1}^\infty \frac1k$ must diverge.

We have $\lim_{x\to \infty}\ln x=\infty$ because $\ln x=\int_1^x(1/t)dt$ is monotonic and $\ln 3^n=n\ln 3>n$ for $n\in \mathbb N.$

For $y>0$ we have $\ln (1+y)|=\int_1^{1+y}(1/t)dt<\int_1^{1+y}1\cdot dt=y. $

Therefore $\sum_{j=2}^n(1/j)>\sum_{j=1}^n \ln (1+1/j)=\sum_{j=1}^n[\;\ln (j+1)-\ln j\;]=\ln (n+1),$ which $\to \infty$ as $n\to \infty.$