Harmonic Series is Divergent

Theorem
$$\sum_{n=1}^\infty \tfrac{1}{n} \ $$ diverges.

Proof
$$\sum_{n=1}^\infty \tfrac{1}{n} = \underbrace{1}_{s_0} + \underbrace{\tfrac{1}{2}+\tfrac{1}{3}}_{s_1} + \underbrace{\tfrac{1}{4}+\tfrac{1}{5}+\tfrac{1}{6}+\tfrac{1}{7}}_{s_2} + ...$$

where $$s_k = \sum_{i=2^k}^{2^{k+1}-1} \tfrac{1}{i}$$

Now $$\forall m  \tfrac{1}{n} \ $$, so each of the summands in a given $$s_k \ $$ are greater than $$2^{-1-k} \ $$. The number of summands in a given $$s_k \ $$ are $$2^{k+1} - 2^k = 2 \times 2^k - 2^k = 2^k \ $$, and so

$$s_k > \tfrac{2^k}{2^{1+k}} = \tfrac{1}{2}$$

Hence the harmonic sum

$$\sum_{n=1}^\infty \tfrac{1}{n} = \sum_{k=0}^\infty \left({s_k}\right) > \sum_{a=1}^\infty \tfrac{1}{2}$$

the last of which diverges.