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Prove the limit $\lim\limits_{n \to \infty}a_n$ exists and evaluate it for a recursive sequence $a_{n+2}=f_{(n\pmod 3)}(a_{n+1},a_n)$ where:

$f_{0}(x,y)={1 \over 2}(x+y)$

$f_{1}(x,y)=\sqrt{xy}$

$f_{2}(x,y)={2 \over x^{-1}+y^{-1}}$

And $a_1$ and $a_2$ are positive real numbers.

I showed it converges - $a_n$ is composed of two suborders, $a_{2n}$ and $a_{2n-1}$, which are both monotone (one increasing and the other decreasing) and bounded, thus each suborder converges. It's easy to show they converge to the same limit, and hence $a_n$ converges. Now, what is the limit?

BONUS: compute $\lim\limits_{a_2 \to \infty}\lim\limits_{n \to \infty} \frac{a_2-a_n}{a_n-a_1}$

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  • $\begingroup$ I don't need $f_3$., yet you're still right. I need n mod 3. $\endgroup$ – Chris May 5 '15 at 7:35
  • $\begingroup$ click on the mod in my comment to know how to write $\mod 3$ properly $\endgroup$ – Elaqqad May 5 '15 at 7:47
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    $\begingroup$ It might help to observe that $0 < x < y \implies (\forall k)~~x < f_k(x,y) < y$ $\endgroup$ – DanielV May 5 '15 at 7:48

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