# Prove a sequence converges using sub-sequences

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 $a^2_{2n+3}-a^2_{2n+4}$?

• 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 – Maciek Jun 30 '15 at 15:42
• 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 – gbox Jun 30 '15 at 15:48
• @Maciek sorry you are right – gbox Jun 30 '15 at 15:50
• @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. – MCT Jun 30 '15 at 15:53
• 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. – 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.
• 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? $a_n\cdot a_n\cdot a_n=L^3$? – gbox Jun 30 '15 at 18:05
• Doesn't $a_{2n+4}\to(x^3-y)^{1/3}$ and similarly for $b_{2n}$? Is the convergence of the 3rd subsequence irrelevant? – user84413 Jun 30 '15 at 19:39
• @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 $x \mapsto x^{1/3}$. – Theo Bendit Jun 30 '15 at 23:27