Is $s_n$ a Cauchy sequence if we only assume that $|s_{n+1} - s_n|\lt \frac{1}{n}$ for all $n\in \Bbb{N}$
My original question was poorly worded, hopefully this make more sense. I get that it is not always convergent. In the theorem, it says for each $\epsilon \gt 0$ there exists N such that n,m $\gt N$ implies
|$s_n - s_m$| $\lt \epsilon$. My main problem is that I do not understand how to relate n and m to n and n+1.
On
We can construct a sequence such that for each $n\geqslant 1$, $s_{n+1}-s_n=1/(2n)$ (for example, define $s_1=0$ and $s_{n+1}=\sum_{i=1}^n1/(2i)$ for $n\geqslant 1$). Then we have $ \lvert s_{n+1}-s_n\rvert=1/(2n)\lt 1/n$ but the sequence $\left(s_n\right)_{n\geqslant 1} $ is not Cauchy, since $s_{2n+1} -s_{n+1}\gt 1/2$ for each $n$.
As stated the given proposition is not correct, since $$ s_n:=\frac1{n^2}, \quad |s_{n+1} - s_n|=\left|\frac{2n +1}{n^2(1+n)^2} \right|<\frac1n,\quad n=1,2,\cdots, $$ and $\left\{s_n\right\}$ is a convergent sequence, thus $\left\{s_n\right\}$ is a Cauchy sequence.