Suppose the series converges to \(H\):
\[H = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + ...\]
Then:
\[H \geq 1 + \frac{1}{2} + \textcolor{red}{\frac{1}{4}} + \frac{1}{4} + \textcolor{red}{\frac{1}{6}} + \frac{1}{6} + \textcolor{red}{\frac{1}{8}} + \frac{1}{8} + \textcolor{red}{\frac{1}{10}} + \frac{1}{10} + ...\]
The terms in red is smaller than the corresponding term of \(H\). Let's group the like terms together:
\[H \geq 1 + \frac{1}{2} + \left(\textcolor{red}{\frac{1}{4}} + \frac{1}{4}\right) + \left(\textcolor{red}{\frac{1}{6}} + \frac{1}{6}\right) + \left(\textcolor{red}{\frac{1}{8}} + \frac{1}{8}\right) + \left(\textcolor{red}{\frac{1}{10}} + \frac{1}{10}\right) + ...\]
Adding the like terms:
\[H \geq 1 + \frac{1}{2} + \left(\textcolor{dimgray}{\frac{1}{2}}\right) + \left(\textcolor{dimgray}{\frac{1}{3}}\right) + \left(\textcolor{dimgray}{\frac{1}{4}}\right) + \left(\textcolor{dimgray}{\frac{1}{5}}\right) + ...\]
Simplify:
\[\begin{aligned} H &\geq 1 + \frac{1}{2} + \textcolor{dimgray}{H - 1} \\ H &\ge \frac{1}{2} + \textcolor{dimgray}H \end{aligned}\]
This contradiction shows that the harmonic series doesn't converge.