Does $(s_{n+1}-s_n)\overset{n\to\infty}{\longrightarrow}0$ in $\mathbb R$ imply that $(s_n)$ converges?

Question

Given a sequence $(s_n)$ in $\mathbb R$ such that $$\lim \limits_{n \to \infty}( s_{n+1}-s_n)=0,$$ I am asked to prove $(s_n)$ converges.

I know all Cauchy sequences converge in $\mathbb R^k$. So I want to prove that $(s_n)$ is Cauchy.

I am stuck as to how to show the given sequence is a Cauchy. Thank you.

Answer 1

It’s not surprising you’re stuck, because the claim is false.

Consider $s_n=\sqrt{n}$. Then $$ \lim_{n\to\infty}(s_{n+1}-s_n)= \lim_{n\to\infty}(\sqrt{n+1}-\sqrt{n})= \lim_{n\to\infty}\frac{1}{\sqrt{n+1}+\sqrt{n}}=0 $$

Answer 2

For any $0 < c < 1$, $s_n =n^c$ satisfies $s_{n+1}-s_n \to 0$ but, obviously, $s_n \to \infty$.

An easy proof is via the mean value theorem. Since $(x^c)' =cx^{c-1} $, then, for some $n \le x \le n+1$, $s_{n+1}-s_n =cx^{c-1} \to 0 $.

Answer 3

Another counter-example is: $s_n=\log n$. Then$$\lim_{n\to\infty}\bigl(\log(n+1)-\log(n)\bigr)=\lim_{n\to\infty}\log\left(1+\frac1n\right)=0.$$