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Let the sequence $\{a_n\}$ be defined as $a_1 = \sqrt 2$ and $a_{n+1} = \sqrt {2+a_n}$.

Show that $a_n \le$ 2 for all $n$, $a_n$ is monotone increasing, and find the limit of $a_n$.

I've been working on this problem all night and every approach I've tried has ended in failure. I keep trying to start by saying that $a_n>2$ and showing a contradiction but have yet to conclude my proof.

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marked as duplicate by YuiTo Cheng, Paul Frost, воитель, José Carlos Santos real-analysis Jul 11 at 12:06

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  • $\begingroup$ i've been working on this problem all night and every approach i've tried has ended in failure. i keep trying to start by saying that $a_n>2$ and showing a contradiction but have yet to conclude my proof $\endgroup$ – david stocker Nov 5 '13 at 17:29
  • $\begingroup$ In the future, please edit your question to include information like that, instead of putting it in a comment. Many people don't bother to read the comments -- they'd miss this important bit of information. $\endgroup$ – Lord_Farin Nov 5 '13 at 17:54
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Hints:

1) Given $a_n$ is monotone increasing and bounded above, what can we say about the convergence of the sequence?

2) Assume the limit is $L$, then it must follow that $L = \sqrt{2 + L}$ (why?). Now you can solve for a positive root.

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  • $\begingroup$ i found the limit no problem, the part i cant do is showing that it is monotone increasing and bounded above. the point of the question is to show convergence $\endgroup$ – david stocker Nov 5 '13 at 17:55
  • $\begingroup$ For the bound, Mathematical Induction seems easy. Check $a_n \le 2 \implies 2+a_n \le 4 \implies \sqrt{2+a_n} \le ...$ you see where this goes? $\endgroup$ – Macavity Nov 5 '13 at 18:02
  • $\begingroup$ "i found the limit no problem," ... why not say that in your statement. Then Macavity would not be giving you hints on how to do it. $\endgroup$ – GEdgar Nov 5 '13 at 19:04
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Complementary to Macavity's answer, the following hints:

  1. For boundedness: $\sqrt x$ is monotone increasing: If $x_1 < x_2$, then $\sqrt{x_1} < \sqrt{x_2}$.

  2. For monotonicity: Since $a_n \le 2$ for all $n$, we have:

    $$2+a_n \ge a_n+a_n = 2a_n \ge a_n\cdot a_n = a_n^2$$

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