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 $H_{2^m}$ satisfies the following inequality:

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

The result follows from the the Comparison Test for Divergence.