# How do I prove this series diverges?

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!

• Okay so I understand you're making use of the fact that it's a decreasing sequence here. Then if I've followed through correctly you'd need to find an $m/n \le 1 - \epsilon$. Now to argue these exist, you take $\epsilon < 1$, you just take any $m \ge n_0$ and then you conclude that the $n$ must exist because of the archimedean property of the real numbers. Is this all correct? – Zeno Jun 20 '17 at 20:54

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
$$\sum_{k = m+1}^n x_k \geqslant (n-m)x_n > \frac{n-m}{n} = 1 - \frac{m}{n} > \frac{1}{2}.$$