Proving sequence $x_n= \sum_{k=1}^{n} \frac{1}{k}$. For $n \geq 2 , \frac{1}{2} \geq \frac{1}{n}$.
so i write $x_n > 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + ... + \frac{1}{2}$.
$x_n > 1 + \frac{n-1}{2}$. Now, $b_n = \frac{n-1}{2} $, $b_n$ diverges $\rightarrow \infty$
The strategy to show this typically involves this comparison of the sum: $$\sum_{k=1}^{2^n}\frac{1}{n}=1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\cdots\geq 1+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+4\cdot\frac{1}{8}+\cdots$$ $$=1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\cdots$$ This strategy was first used by Nicole Oresme around 1350.
Alternatively, on a similar process on every even term after $x_2$, you can show the sum is greater than or equal to $\frac{1}{2}+x_n$, so the sum diverges.