Proof that $\sum\limits_{i=1}^n \left(\dfrac{1}{i}\right)$ is divergent

 (1)  Assume that $\sum\limits_{i=1}^n \left(\dfrac{1}{i}\right)$ is convergent so that there exists $H$ such that:

$$H = \sum\limits_{i=1}^n \left(\dfrac{1}{i}\right) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots$$

(2)  $H \ge 1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{8} + \dfrac{1}{8} + \dots =$

$$= 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots = \frac{1}{2} + H$$

(3)  Since we have reached a contradiction, we can reject the assumption and conclude that the sequence is divergent.