Consider a decreasing sequence $(x_n)$ in $\Bbb{R}_0$. There are an infinite amount of $n \in \Bbb{N}_0$ for which $1/n < x_n$. Prove the series $\sum x_n$ diverges.
On one hand, I considered proving the sequence $(x_n)$ does not converge to $0$. However, unless I'm mistaken, this is not necessarily true. My next attempt was trying to show the series is not Cauchy. In other words:
Find an $\epsilon > 0$ such that for every $n_0 \in \Bbb{N}_0$ an $m,n > n_0$ exists for which $\epsilon \le |\sum_{m}^{n} x_k|$.
I figured I'd try to pick one of those $x_n > 1/n$, and a certain $N$ amount of preceding elements to reach the conclusion that $\epsilon \le \frac{N}{n} \le |\sum_{m}^{n} x_k|$. However at this point, I'm completely lost on how to prove these numbers $N$ and $n$ exist. Am I headed in the right direction? And if so, how do I finish the proof? Thanks in advance!
Yes, you're heading very much in the right direction. The point to finish it is to fix the lower limit of the sum (arbitrarily), and then use that there are arbitrarily large $n$ with $x_n > 1/n$. So if, for an arbitrary $m$ we choose an $n > 2m$ with $x_n > 1/n$, we find
$$\sum_{k = m+1}^n x_k \geqslant (n-m)x_n > \frac{n-m}{n} = 1 - \frac{m}{n} > \frac{1}{2}.$$