Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

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 $2^{2(1+1/2+1/4+\dots)}=16$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.