Harmonic Series is Divergent

Theorem

 * $\displaystyle \zeta(1) = \sum_{n=1}^\infty \frac{1}{n} \ $ diverges.

Proof

 * $\displaystyle \sum_{n=1}^\infty \frac{1}{n} = \underbrace{1}_{s_0} + \underbrace{\frac{1}{2}+\frac{1}{3}}_{s_1} + \underbrace{\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}}_{s_2} + \cdots$

where $\displaystyle s_k = \sum_{i=2^k}^{2^{k+1}-1} \frac{1}{i}$.

Now $\forall m < n: \dfrac{1}{m} > \dfrac{1}{n} \ $, so each of the summands in a given $s_k \ $ is greater than $\dfrac 1 {2^{k+1}} \ $.

The number of summands in a given $s_k \ $ is $2^{k+1} - 2^k = 2 \times 2^k - 2^k = 2^k \ $, and so:
 * $s_k > \dfrac{2^k}{2^{1+k}} = \dfrac{1}{2}$

Hence the harmonic sum:
 * $\displaystyle \sum_{n=1}^\infty \frac{1}{n} = \sum_{k=0}^\infty \left({s_k}\right) > \sum_{a=1}^\infty \frac{1}{2}$

the last of which diverges.

Also see

 * Harmonic numbers
 * Riemann Zeta Function