I'm thinking about the following statement:
Let $a_n$ be any real sequence, if $\lim_{n \to \infty} |a_{n+1}-a_n|=0$ then $a_n$ converges.
What comes in mind is to use Cauchy's criteria but it seems like what I'm trying to prove here is weaker, which makes me think that maybe it's not true (although it sounds true).
EDIT:
How to prove that $a_n$ converges after strengthening the assumption, that is, we assume that:
$$\lim_{n\to \infty}2^n|a_{n+1}-a_n|=L>0$$
It then becomes obvious that
$$\frac{L-\varepsilon}{2^n}<|a_{n+1}-a_n|<\frac{L+\varepsilon}{2^n}$$
and then
$$|a_{n+1}|<|a_n|+\frac{L+\varepsilon}{2^n} $$
so that
$$|a_{n+1}-a_n|<\left||a_n|+\frac{L+\varepsilon}{2^n} -a_n \right|$$
However, I can't seem to make progress.
Best Answer
The statement is false. Take, for example, $$ a_n=\sum_{i=1}^n\frac1i $$ for $n\ge1$. Then $|a_{n+1}-a_n|=1/(n+1)\to0$ as $n\to\infty$, but the sequence $a_n$ does not converge (see Harmonic series).