The solution requires a little bit of creativity, but once you know the trick, you'll see that it's actually pretty easy. I recommend mastering it since it always come up now and there, be it at a math interview or in your university journey.

The trick is to visualise each \(\frac1k\) in \(\sum_{k=1}^n\frac1k \) as the area of a rectangle of dimension \(1 \times \frac1k\) and to consider the function \(x \rightarrow \frac1x\):

As you can see \(\sum_{k=1}^\infty\frac1k \) simply corresponds to the sum of all the rectangles' area, which is strictly greater than the area under the curve \(x \rightarrow \frac1x \) for \(x \geq 1 \). Hence, if you can show that the area under that curve is unbounded, you win.

The area under the curve \(x \rightarrow \frac1x \) from \(x=1\) to \(x=N\) is simply: $$ \int_1^N {\frac1x} \, dx = \bigg[\log(x)\bigg]_1^N $$ $$ = \log(N) - \log(1)$$ $$ = \log(N) $$ which tends to \(\infty\) as \(N \rightarrow \infty\). Hence \(\sum_{k=1}^n\frac1k \rightarrow \infty \text{ as } n \rightarrow \infty \).