If $\lim \limits_{n \to \infty\ } (a_{n+1}-a_n)=0$ and $(a_{4n})_{n\ge1}$ converges decide whether or not $(a_n)$ converges.

Question

If $\lim \limits_{n \to \infty\ } (a_{n+1}-a_n)=0$ and $(a_{4n})_{n\ge1}$ converges decide whether or not $(a_n)$ converges.

My attempt:

I think it isn't enough to $a_n$ converges.

I know $\lim \limits_{n \to \infty \ }( a_{4n+1}-a_{4n})=0$, and because $(a_{4n+1}−a_{4n})$ is a subsequence of $(a_{n+1}-a_n)$ it also converges to 0, so $\lim \limits_{n \to \infty\ } a_{4n+1}=\lim \limits_{n \to \infty\ } a_{4n} $

But I don't know how how to finish this proof. Could someone give me some hints?

Answer 1

Write $\ell = \lim_{n\to\infty} a_{4n}$. For each $\epsilon > 0$,

• Pick $N_1$ such that $n \geq N_1$ implies $|a_{n+1} - a_n| < \epsilon/5$.
• Pick $N_2$ such that $4k \geq N_2$ implies $|a_{4k} - \ell| < \epsilon/5$.

Write $N = \max\{N_1, N_2\}$. Then for each $n \geq N$, with $n = 4k+r$ with $r \in \{0,1,2,3\}$,

$$ |a_n - \ell| \leq \left( \sum_{j=r}^{3} |a_{4k+j+1} - a_{4k+j}| \right) + |a_{4(k+1)} - \ell| < \epsilon. $$

Therefore the $\epsilon$-$N$ definition of $a_n \to \ell$ is established and hence $a_n \to \ell$.

Answer 2

By the same technique you used, $(a_{4n+2} - a_{4n+1})$ is a subsequence of $(a_{n+1} - a_n)$, so it has to converge to $0$ as well. Does that help?

Answer 3

Hints: let $j,k\in \mathbb N$ and write $j=4j'+r,k=4k'+s$ with $0\leq r,s <4$. Using the fact that $a_{n+1}-a_n \to 0$ show that $a_{n+2}-a_n \to 0$ and $a_{n+3}-a_n \to 0$. Conclude that $a_j-a_{4j'} \to 0$ and $a_k-a_{4k'} \to 0$. Now can you complete the proof?