The harmonic series can be shown to diverge by a simple argument involving grouping terms. Write $$ \sum_{n=1}^{2^M} \frac{1}{n} = \sum_{m=1}^M \sum_{n=2^{m-1}+1}^{2^m} \frac{1}{n}. $$ Since $1/n \ge 1/N$ when $n \le N$ we have $$ \sum_{n=2^{m-1}+1}^{2^m} \frac{1}{n} \ge \sum_{n=2^{m-1}+1}^{2^m} 2^{-m} = (2^m - 2^{m-1}) 2^{-m} = \frac{1}{2} $$ Hence, $$ \sum_{n=1}^{2^M} \frac{1}{n} \ge \frac{M}{2} $$ so the series diverges in the limit $M \to \infty$