8
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ Absolutely correct. $\endgroup$ – Alex M. Apr 21 '15 at 13:31
  • 1
    $\begingroup$ Isn't it enough to ask for $\lim_{n\to \infty} a_{n+1}-a_n=0$? $\endgroup$ – Alberto Debernardi Apr 21 '15 at 13:34
  • 3
    $\begingroup$ @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? $\endgroup$ – Ilham Apr 21 '15 at 13:45
  • 1
    $\begingroup$ Actually I answered a question some days ago which was very similar to what you've posted: math.stackexchange.com/questions/1239056/… . 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. $\endgroup$ – Alberto Debernardi Apr 21 '15 at 14:04
  • 1
    $\begingroup$ @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? $\endgroup$ – user233191 Apr 21 '15 at 14:43
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.