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Let $(a_n)$ be a sequence such that $\lim_{n\to\infty} (a_{n+1}-a_n)=0$ and $|a_{n+2}-a_n|<\frac{1}{2^n}$ for all $n$. I have to decide whether or not $(a_n)$ converges.

My attempt: I think it converges. Let $b_n=a_{2n}, c_n=a_{2n-1}$. Then:

$$|b_{n+1}-b_n|=|a_{2n+2}-a_{2n}|<\frac{1}{2^{2n}}$$

$$|c_{n+1}-c_n|=|a_{2n+1}-a_{2n-1}|<\frac{1}{2^{2n-1}}$$

Thus $(b_n)$ and $(c_n)$ are Cauchy (proven in another question) and converge. Because $(a_{2n}-a_{2n-1})$ is a subsequence of $(a_{n+1}-a_n)$ it also converges to $0$. Thus $$\lim_{n\to\infty} (b_n-c_n)=\lim_{n\to\infty} (a_{2n}-a_{2n-1})=0$$

or

$$\lim_{n\to\infty} b_n=\lim_{n\to\infty}c_n$$

Because the subsequences $(b_n)$ and $(c_n)$ cover the sequence $(a_n)$ and because they converge to the same point, $(a_n)$ converges.

Is it correct? What do you think?

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user233191
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    Absolutely correct. – Alex M. Apr 21 '15 at 13:31
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    Isn't it enough to ask for $\lim_{n\to \infty} a_{n+1}-a_n=0$? – Alberto Debernardi Apr 21 '15 at 13:34
  • I would think so, it should be enough to prove it is a Cauchy sequence. – yellon Apr 21 '15 at 13:36
  • Roughly speaking the limit condition is telling us that the terms of the sequence accumulate at some point, that they are "all together" for $n$ big enough. So the point where they accumulate should actually be the limit of the sequence. – Alberto Debernardi Apr 21 '15 at 13:41
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    @albertodebernardi Is it really enough? Doesn't the sequence ln n have the property that the difference between terms tends to 0, but isn't convergent? – Ilham Apr 21 '15 at 13:45
  • @Ilham Correct, thank you for providing the counterexample. So an arising question now is, which is the weakest additional condition we should add in order to have convergence? – Alberto Debernardi Apr 21 '15 at 13:48
  • @albertodebernardi this might be a weak enough. So the difference must not only converge to 0, but must converge fast enough. Http://math.stackexchange.com/questions/182830/proving-that-a-sequence-is-cauchy – Ilham Apr 21 '15 at 14:01
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    Actually I answered a question some days ago which was very similar to what you've posted: http://math.stackexchange.com/questions/1239056/prove-that-if-x-n1-x-n-leq-frac1n2-for-all-terms-of-a-sequence-t/1239066#1239066 . In view of the answer, I would conjecture that if $|a_n-a_{n+1}|\leq c_n$ and $c_n$ is a summable sequence, then $a_n$ is Cauchy and therefore convergent. – Alberto Debernardi Apr 21 '15 at 14:04
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    @AlbertoDebernardi - if I understand correctly, the condition you presented is needed to prove that $(b_n)$ and $(c_n)$ are Cauchy and therefore convergent (which leads to the convergence of $(a_n)$), correct? And the given inequality is just a special case of your condition, right? – user233191 Apr 21 '15 at 14:43
  • With that condition you can directly prove the convergence of $(a_n)$ without looking into the subsequences. – Alberto Debernardi Apr 21 '15 at 17:44

1 Answers1

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Here you have a sufficient condition (from the discussion in the comments) for your sequence to be convergent: Suppose that $|a_n-a_{n+1}|\leq c_n$, where $\{c_n\}$ is such that $$ \sum_{n=1}^{\infty}c_n<\infty. $$ We define the sequence of positive numbers $C_n=\sum_{k=1}^nc_n$, which is increasing. Since it converges, it is also Cauchy. Now we prove that $a_n$ is Cauchy: let $m> n$ $$ |a_n-a_m|=|a_n-a_{n+1}+a_{n+1}-\cdots -a_m|\leq \sum_{k=n}^{m-1} |a_k-a_{k+1}|\leq \sum_{k=n}^{m-1} c_k = C_m-C_{n-1} \to0 $$ as $m,n\to \infty$.

Sorry for off-topic but the discussion in the comments was quite rich in my opinion, worth to be completed.

EDIT: Actually we only need the summability of the sequence $\{|a_n-a_{n+1}|\}$. Such sequences $\{a_n\}$ are called of bounded variation. So, supposing $$ \sum_{n=1}^{\infty}|a_n-a_{n+1}|<\infty, $$ then it suffices to take $c_n=|a_n-a_{n+1}|$ in the previous argument.

Alberto Debernardi
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