Harmonic Series is Divergent/Proof 1

Proof

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

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

From Ordering of Reciprocals:
 * $\forall m, n \in \N_{>0}: m < n: \dfrac 1 m > \dfrac 1 n$

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:


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

Hence the harmonic sum satisfies the following inequality:


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

The rightmost expression diverges, from the $n$th term test.

The result follows from the the Comparison Test for Divergence.