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Find the limit of the following recurrence relation: $$a_{n+1}=\sqrt{7-(-1)^na_n}, n\geq 0$$ with $a_0=0$.

I have thought that we can transform the relation to the following: $$a_{n+1}^2+(-1)^na_n=7$$ but I cannot take it further!!! Any help? (I also don't know how to prove the convergence of such a sequence)

Edit 1: I think that it is not a duplicate, because my relation has also the $(-1)^n$ part.

  • possible duplicate of Limit of this recursive sequence and convergence –  Sep 12 '15 at 18:00
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    Have you calculated the early part of the sequence? Seems clear to me that the odd and even subsequence converge to different limits, and hence the entire sequence diverges. Probably that is not exceedingly difficult to prove either. – Jyrki Lahtonen Sep 12 '15 at 18:01
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    Additionally, these two limit points are 2 (for odd $n$) and 3 (for even $n$). – Tintarn Sep 12 '15 at 18:02
  • @Jason: What exactly do you mean by "the limit"? As pointed out by Jyrki Lahtonen above, there exists no limit of the sequence. Rather, there exist two limit points one of which the sequence $a_{2n}$ converges to (it's 3) and one where the sequence $a_{2n+1}$ converges to (it's 2). – Tintarn Sep 12 '15 at 18:17

2 Answers2

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Hint. Note that the subsequences $(a_{2n})$ and $(a_{2n+1})$ have different limits.

Also, $$ b_{n+1}=a_{2n+2}=\sqrt{7-a_{2n+1}}=\sqrt{7-\sqrt{7+a_{2n}}}=\sqrt{7-\sqrt{7+b_n}} $$ Similarly $$ c_{n+1}=a_{2n+1}=\sqrt{7+\sqrt{7-a_{2n-1}}}=\sqrt{7+\sqrt{7-c_{n-1}}} $$ Clearly, $(b_{2n})$, $(c_{2n})$, $(b_{2n+1})$, $(c_{2n+1})$ are monotonic, bounded and hence convergent.

For $(b_{n})$, if $x$ is the limit, then $$ x=\sqrt{7-\sqrt{7+x}} $$ or $(7-x^2)^2=7-x$...

Yiorgos S. Smyrlis
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    May I ask why $(b_n)$ and $(c_n)$ are monotonic? – eudes Sep 12 '15 at 20:04
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    Actually, no. I don't agree they are. Although they do seem to converge quickly and the subsequences $(b_{2k})$, $(b_{2k+1})$, $(c_{2k+2})$, $(c_{2k+1})$ seem to be monotonic. – eudes Sep 12 '15 at 20:40
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One can show that the limit does not exist by contradiction. Suppose $\lim_na_n$ exists. Then we must have that $x=:\lim_na_{2n}$ and $y:=\lim_na_{2n+1}$ both exists. We show that $x\neq y$ and thus have a contradiction.

By the recursive relation, one has $$ a_{2n+1}=\sqrt{7-\sqrt{7+a_{2n-1}}},\quad a_{2n}=\sqrt{7+\sqrt{7-a_{2n-2}}} $$ which implies by taking the limit and the continuity of square root that $$ y=\sqrt{7-\sqrt{7+y}}\qquad(1)\\ x=\sqrt{7+\sqrt{7-x}}\qquad(2) $$ One can check that solutions to (1) and (2) can never be the same: enter image description here enter image description here