# Show $a_{n+1}=\sqrt{2+a_n},a_1=\sqrt2$ is monotone increasing and bounded by $2$; what is the limit? [duplicate]

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.

## marked as duplicate by YuiTo Cheng, Paul Frost, воитель, José Carlos Santos real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 11 at 12:06

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• 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 – david stocker Nov 5 '13 at 17:29
• 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. – Lord_Farin Nov 5 '13 at 17:54
• – YuiTo Cheng Jul 11 at 8:01

## 2 Answers

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.

• 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 – david stocker Nov 5 '13 at 17:55
• 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? – Macavity Nov 5 '13 at 18:02
• "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. – GEdgar Nov 5 '13 at 19:04

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:

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