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Let $a_{n+1}=\sqrt {(a_n+a_{n-1})/2}$ and $a_0=a_1=2$, how to prove convergence of the product $a_0 a_1 a_2 a_3...a_\infty$, and possibly find its value?

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1 Answer 1

up vote 2 down vote accepted

Show by induction that $a_n$ is decreasing:

$a_{n}+a_{n-1}\le a_{n-1}+a_{n-2}$ implies $(a_{n}+a_{n-1})/2\le (a_{n-1}+a_{n-2})/2$ implies $a_{n+1}\le a_{n}$.

So, in particular, $a_{n} \le \sqrt{a_{n-2}} \le 2^{2^{-\lfloor n/2 \rfloor}}$ and the product is bounded above by $2^{2(1+1/2+1/4+\dots)}=16$.

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