Let there be a sequence $a_n$
The following sub-sequences converge: $a_{n^3},a^3_{2n+3}-a^3_{2n+4},a^2_{2n+3}-a^2_{2n+4},a_{2n+15}$
Prove: $a_n$ converges

I think it has something to do with binomial due to the given sub-sequences

for example: $(a_{2n+3}-a_{2n+4})\cdot (a_{2n+3}+a_{2n+4})=a^2_{2n+3}-a^2_{2n+4}$ so can I say something about the components of $a^2_{2n+3}-a^2_{2n+4}$?

  • $\begingroup$ If $a_{2n+15}$ converges, then you know that the odd elements of the sequence are also convergent. Then use the other information to infer things about the even terms $\endgroup$ – Maciek Jun 30 '15 at 15:42
  • $\begingroup$ So I can use $a_n^3$ to say something about the even places? Just like $a_{2n+15}$ it not true for all odds, just those with a gap of 15 $\endgroup$ – gbox Jun 30 '15 at 15:48
  • $\begingroup$ @Maciek sorry you are right $\endgroup$ – gbox Jun 30 '15 at 15:50
  • $\begingroup$ @Maciek I don't know what you have in mind, but for a sequence the convergence of odd and even elements doesn't imply convergence of the entire sequence. It could be the odds converging to $7$ and the evens converging to $3$, for example. $\endgroup$ – MCT Jun 30 '15 at 15:53
  • 1
    $\begingroup$ I only ask because the answer seems perfectly correct and elegant, I fear that the question is not as good as the answer and you might be willing to lower the quality of your answer. $\endgroup$ – Conrado Costa Jul 6 '15 at 21:29

The odds converge, since $a_{2n+15}$ converges. Let their limit be $x$ and let the limit of $a_{2n+3}^3 - a_{2n+4}^3$ be $y$. Then, by algebra of limits, $$a_{2n+4} = (a_{2n+3}^3 - (a_{2n+3}^3 - a_{2n+4}^3))^{1/3} \rightarrow (x^3 - y)^{1/3}.$$ Thus the evens also converge. Let $b_n = a_{n^3}$, and $z$ be its limit. Consider the subsequences $b_{2n}$ and $b_{2n+1}$. These are also subsequences of the odd and even terms of $a_n$, hence they converge respectively to $x$ and $(x^3 - y)^{1/3}$. But, they are subsequences of $b_n$, so they also converge to $z$. By uniqueness of limits, $x = z = (x^3 - y)^{1/3}$. So, the odd and even terms of $a_n$ converge to the same limit, so the entire sequence converges to that limit.

  • $\begingroup$ Can we say that $a^3_{2n+3}$ as a limit? we know that every odd subsequence of $a_n$ when $n>15$ have a limit, can we use arithmetic of limits on 3 limits? $a_n\cdot a_n\cdot a_n=L^3$? $\endgroup$ – gbox Jun 30 '15 at 18:05
  • 2
    $\begingroup$ Doesn't $a_{2n+4}\to(x^3-y)^{1/3}$ and similarly for $b_{2n}$? Is the convergence of the 3rd subsequence irrelevant? $\endgroup$ – user84413 Jun 30 '15 at 19:39
  • 1
    $\begingroup$ Fixed! Thanks user84413. $\endgroup$ – Theo Bendit Jun 30 '15 at 23:22
  • 3
    $\begingroup$ @gbox: Yes. If $a_n$ converges to $L$, then $a_n \cdot a_n \rightarrow L \cdot L$, and $a_n \cdot (a_n \cdot a_n) \rightarrow L \cdot (L \cdot L)$. More generally, if $f$ is a real function that is continuous at $L$, then $f(a_n) \rightarrow f(L)$. I used the continuity of both the functions $x \mapsto x^3$ and $x \mapsto x^{1/3}$. $\endgroup$ – Theo Bendit Jun 30 '15 at 23:27

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.