TITLE: Show $\sum_\limits{k=1}^{\infty}1/k$ does not converge.
QUESTION [6 upvotes]: Show $$\sum_\limits{k=1}^\infty \frac 1 k$$ does not converge.


Attempt:
Let $s_n=\sum_\limits{k=1}^{n}1/k$, and let $\epsilon=1/2$. For all $N\in\mathbb{N}$, we have $$\left|s_{2n}-s_n\right|=\left|\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}\right|\geq1/2,\qquad\text{for all $n\geq N$}$$ Hence, $\{s_n\}_{n=1}^{\infty}$ is not a Cauchy sequence.
Since $\{s_n\}_{k=1}^{\infty}$ is not a Cauchy sequence which implies $\{s_n\}$ doesn't converge, we have $$\lim\limits_{n\rightarrow\infty}s_n=\lim\limits_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{k}=\infty$$ Therefore, the infinite series $\sum_\limits{k=1}^{\infty}\frac{1}{k}$ does not converge.

I am not sure this is valid or not because I use contradiction to do these kind problem.

REPLY [2 votes]: Your proof is fine, it just need a minor fix: since the sequence $\{s_n\}_{n\geq 1}$ $\color{red}{\text{is increasing}}$ and it is not a Cauchy sequence by the shown inequality, $\lim_{n\to +\infty}s_n = +\infty$.
You need the red part, since a sequence that is not a Cauchy sequence may also be an oscillating sequence, like in the case $a_n = \sin(n)$ or $a_n = 2+\cos(n)$.
The Cauchy condensation test is the usual way for proving that the harmonic series is divergent, but there are plenty of alternatives. For instance, since $f(x)=\frac{1}{x}$ is a convex function on $\mathbb{R}^+$, the Hermite-Hadamard inequality gives:
$$ \frac{1}{2}\left(\frac{1}{1}+\frac{1}{n}\right)+\left(\frac{1}{2}+\ldots+\frac{1}{n-1}\right)\geq \int_{1}^{n}\frac{dx}{x}$$
from which it follows that:
$$ H_n=\sum_{k=1}^{n}\frac{1}{k} \geq \log(n)+\frac{1}{2}. $$