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June 6th, 2015, 05:33 AM  #1 
Newbie Joined: Jun 2015 From: On the Moon. Posts: 1 Thanks: 0  Examine the convergence of a sequence $\{a_{n}\}$ which is given by $a_{1}=a>0,a_{2}=
I used inequality between arithmetic and geometric means to show that a sequence $\{a_{n}\}$ is bounded: $$a_{n+2}=\sqrt{a_{n+1}a_{n}}\le \frac{a_{n+1}+a_{n}}{2}$$ Solving this, I get quadratic inequality $$a^2_{n+1}2a_{n+1}a_{n}+a^2_{n}\ge 0$$ which gets me to $$0 \le a_{n}\le 1$$ thus, sequence is bounded. Is this right? How to show that the sequence is monotonic? We don't know if $$a_{n+2}>a_{n+1}$$ or $$a_{n+2}<a_{n+1}$$ Do we need to use induction? 
June 6th, 2015, 06:43 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra 
It's not monotonic because $a \le \sqrt{ab} \le b$ when $a \le b$, but the subsequences $\{\ldots, a_{n2},a_n,a_{n+2},\ldots\}$ are monotonic. You should concentrate on proving that both subsequences converge to the same value. You can probably guess the value the converge to, and this can be useful.

June 7th, 2015, 09:14 PM  #3 
Newbie Joined: Aug 2014 From: Kalispell MT USA Posts: 20 Thanks: 0 
An can = 2 when 0=A over N=√ of 4 But greater then 1 is N over 0 

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$a1a>0, $an$, calculus, convergence, examine, sequence 
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