$$\sum_{n=1}^{\infty} \frac{1}{\sqrt n}$$
Obviously, this is the p-series (hyperharmonic series) with $p=\frac{1}{2} < 1$ which means that it is divergent, so that's what I'm going to try to prove with the following criterion, which is the negation of the Cauchy's convergence test:
The series $\sum_{n=1}^{\infty} a_n$ diverge iff $\exists\; \varepsilon > 0 \space \forall\; N \in \mathbb{N} \space \exists\; p,n \in \mathbb N \colon \require{enclose} \enclose{horizontalstrike}{p \geq n} n \geq N$ such that $|a_{n+1} + \dots + a_{n+p}| \geq \varepsilon$
Working backwards as usual, I start with $$\Bigl| \frac{1}{\sqrt{n+1}} + \dots + \frac{1}{\sqrt{n+p}} \Bigr| = \frac{1}{\sqrt{n+1}} + \dots + \frac{1}{\sqrt{n+p}} \geq \frac{1}{\sqrt{n+1}},\; \forall\; n$$
Now for a choice of $\varepsilon = 1/2$, we can choose $n=1$ and let $p \in \mathbb N$ be fixed. Then $\frac{1}{\sqrt{1+1}} = \frac{1}{\sqrt{2}} > \frac{1}{2}$.
I'm not sure if this is on the right track and if I should have done anything with $N$, perhaps just say "let N > 1""let $N = n - 1$"?