Show that sequence has limit We know that(1) $\displaystyle \lim_{n \to \infty}{(a_{n+1}-a_n)}=0$  and (2) $\displaystyle |a_{3m}-a_{3n}|<\varepsilon$ show that $a_n$ converge and explain why it's not sufficient to converge when $a_n$ satisfy (1) but not (2) and (2) but not (1)
 A: Hint: Prove that it is a Cauchy sequence.
There are 9 cases for 
$
|a_n - a_m|
$:


*

*$|a_{3p} - a_{3q}|$

*$|a_{3p} - a_{3q-1}|$

*$|a_{3p} - a_{3q-2}|$

*$|a_{3p-1} - a_{3q}|$

*$|a_{3p-1} - a_{3q-1}|$

*$|a_{3p-1} - a_{3q-2}|$

*$|a_{3p-2} - a_{3q}|$

*$|a_{3p-2} - a_{3q-1}|$

*$|a_{3p-2} - a_{3q-2}|$


Start assuming that
$$
\min(p-2, q-2, n) > N\implies \max(|a_{3p} - a_{3q}|, |a_{n} - a_{n+1}|) < \frac\epsilon5
$$
A: Proof. Suppose (1)(2) holds. (2) implies that $\exists A\in\mathbb{R}$ such that $$\lim_{m\to \infty}a_{3m}=A.$$ Fix $\epsilon>0$. Let $N_1\in \mathbb{N}$ be such that if $m>N_1$, then $|a_{3m}-A|\le \epsilon/3$. Let $N_2\in \mathbb{N}$ be such that if $n>N_2$, then $|a_{n+1}-a_n|\le \epsilon/3$. Then for any $n>3N_1+3N_2$ we have
(i) if $n=3m$, then $m>N_1$, thus $|a_n-A|\le \epsilon/3$;
(ii) if $n=3m\pm1$, then $m>N_1$, thus $|a_n-A|\le |a_{3m}-a_{3m\pm 1}|+|a_{3m}-A|\le  2\epsilon/3$. 
So we always have $$|a_n-A|\le 2\epsilon/3<\epsilon$$ whenever $n\ge 3N_1+3N_2$. Done!
Counterexamples. 
(a) Let $a_n=\sum_{k=1}^n \frac 1k$, then (1) holds, (2) fails. 
(b) Let $a_{3m}=0$ for all $m$, $a_{3m\pm 1}=1$. Then (2) holds, (1) fails.
Both sequence have no limit.
