Prove the convergence My task is to prove, that if this both conditions hold$$\lim_{n\rightarrow\infty}(a_{n+1}-a_n)=0$$ $$\forall{\varepsilon >0} \exists{N\in\mathbb{N}}\forall{n,m>N}: |a_{3m}-a_{3n}|\leq\varepsilon$$then $a_{n}$ converges, and show examples, that none of above is alone sufficient for $a_{n}$ to converge.
I am afraid I don't understand second condition good enough to do anything. It seems similiar to Cauchy's condition of limit, but if it is so, then $a_{3n}$ is a kind of limit for $a_{3m}$. But what does it mean that subseries are bounded by themselves? What about examples? I would greatly appreciate any help and I look forward to it. Thank you  in advance
 A: Hint: any $n \in \mathbb{N}$ can be written either as $3k$, or $3k+1$, or $3k+2$ for some $k \in \mathbb{N}$. The sequence $\{a_{3k}\}_k$ converges. Moreover
$$
|a_{3k+1}-a_{3k}| \leq \varepsilon
$$
and
$$
|a_{3k+2}-a_{3k}| \leq |a_{3k+1+1}-a_{3k+1}| + |a_{3k+1}-a_{3k}| \leq 2 \varepsilon
$$
when $k \gg 1$.
A: Hint: You'll want to show that the sequence is Cauchy. Use the fact that every number can be written as $3k$, $3k-1$, or $3k+1$ for some $k$, and use a triangle inequality argument with each term eventually bounded above by $\epsilon/3$.
For a counterexample satisfying the first condition only, consider $a_n=\log n$.
For a counterexample satisfying the second condition only, $a_n=\begin{cases}1 & \text{if }n\text{ is an integer multiple of }3\\0 & \text{otherwise.}\end{cases}$
A: The second condition says that the subsequence $(a_{3n})_{n\in\mathbb N}$ is Cauchy, hence converges to some limit $a$. Together with the first condition, the sequence itself converges: Let $\epsilon>0$ be given. By the convergence of $a_{3n}$, there exists $N_1$ such that $|a_{3n}-a|<\frac\epsilon2$ for all $n>N_1$. By the first condition, there exists $N_2$ such that $|a_{n+1}-a_n|<\frac\epsilon2$ for all $n>N_2$. Now let $N=\max\{3N_1+2, N_1\}$. If $n>N$, then either $n=3k$ or $n=3k+1$ or $n=3k+2$ with $k>N_1$. If $n=3k$, then $$|a_n-a|=|a_{3k}-a|<\frac\epsilon2.$$ If $n=3k+1$, then $$|a_n-a|\le |a_{3k+1}-a_{3k}|+|a_{3k}-a|<\frac\epsilon2+\frac\epsilon2=\epsilon.$$
If $n=3k+2$, then 
$$|a_n-a|\le |a_{3k+2}-a_{3(k+1)}|+|a_{3(k+1)}-a|<\frac\epsilon2+\frac\epsilon2=\epsilon. $$
At any rate $|a_n-a|<\epsilon$ for all $n>N$.

A sequence where the first condition holds, but that does not converge, is given by $a_n =\sqrt n$.
A sequence where the second condition holds, but the sequence does not converge, is given by $a_n= n\bmod 3$.
A: Yes, you're on the right track: try to prove that $(a_n)$ is Cauchy. For arbitrary $n,m$, say if $n=3k+r$ and $m=3l+s$ where $r,s\in\{0,1,2\}$, then
$$|a_m-a_n|\le \overbrace{|a_m-a_{m-1}| + \ldots }^{r\text{ pieces}}+|a_{3k}-a_{3l}| + \overbrace{\ldots +|a_m-a_{m-1}| }^{s\text{ pieces}} < 2\frac\varepsilon5+\frac\varepsilon5+2\frac\varepsilon5 $$
if $n,m>N_0$ for some $N_0$, being the maximum of $N_0$'s belonging to $\varepsilon/5$..
