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

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.
