Why does the condition $|a_{n+1}-a_n|\le1/n$ not imply the sequence is Cauchy?

Question

While going through this condition I noticed that the sequence $a_n=\sum_{n=1}^{\infty}1/n$ satisfies $|a_{n+1}-a_n|\le1/n$ $\forall n\in$ $\mathbb{N}$. However $a_n$ is not a Cauchy sequence, and with this I had established a counter-example for the condition. But ultimately I failed to understand why such a thing happened even though the difference of the terms were bounded by $1/n$, and $1/n$ is a sequence whose terms condense to a point as $n$ tends to $+\infty$(essentially the idea behind Cauchy Sequences). Also how can I check if the following conditions imply "Cauchyness": 1.$|a_{n+1}-a_n|\le(n+1)^n/n^{3/2+n}$, 2.$|a_{n+1}-a_n|\le (n\log\ n)/e^n$

Answer 1

Cauchy sequences are about $|a_m-a_n|$ being arbitrarily small if $m$ and $n$ are both large enough

But in this case you have $a_{2n}-a_n =\sum\limits_{j=n+1}^{2n}\frac1j \ge n \frac1{2n} = \frac12$, which does not get arbitrary small.