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June 6th, 2015, 05:33 AM   #1
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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?
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June 6th, 2015, 06:43 AM   #2
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It's not monotonic because $a \le \sqrt{ab} \le b$ when $a \le b$, but the subsequences $\{\ldots, a_{n-2},a_n,a_{n+2},\ldots\}$ 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.
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June 7th, 2015, 09:14 PM   #3
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An can = 2 when 0=A over N=√ of 4
But greater then 1 is N over 0
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