# Show that $\sqrt{2+\sqrt{2+\sqrt{2…}}}$ converges to 2 [duplicate]

Consider the sequence defined by $a_1 = \sqrt{2}$, $a_2 = \sqrt{2 + \sqrt{2}}$, so that in general, $a_n = \sqrt{2 + a_{n - 1}}$ for $n > 1$. I know 2 is an upper bound of this sequence (I proved this by induction). Is there a way to show that this sequence converges to 2? What I think is that the key step is to prove 2 is the least upper bound of this sequence. But how?

## marked as duplicate by Najib Idrissi, Tunk-Fey, Martin Sleziak, Jack D'Aurizio, Davide Giraudo 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(); } ); }); }); Aug 31 '14 at 10:27

• math.stackexchange.com/questions/849274/… – Bumblebee Aug 24 '14 at 16:09
• $a_n$ converges to $2$ because it's monotone increasing and bounded by $2$. With induction you get $\sqrt{2+a_{n+1}} = \sqrt{2+\sqrt{2+a_n}} \leq \sqrt{2+2} = 2$ – monoid Aug 24 '14 at 16:13

Let $x = \sqrt {2 + \sqrt {2 + \sqrt {2 + \cdots}}}$. Then, note that $$x^2 = 2 + \sqrt {2 + \sqrt {2 + \cdots}} = 2 + x \implies x^2 - x - 2 = 0.$$Note that the two solutions to this equation are $x=2$ and $x=-1$, but since this square root cannot be negative, it must be $2$.

• You haven't proven convergence. – Najib Idrissi Nov 7 '13 at 16:33
• If it didn't converge, I wouldn't get a valid solution. – Ahaan S. Rungta Nov 7 '13 at 16:34
• Really? Take the sequence $a_0 = 1, a_{n+1} = 2 a_n$. Then if the limit exists, it must satisfy $x = 2x$, and therefore $x = 0$. Do you think the sequence converges to 0? – Najib Idrissi Nov 7 '13 at 16:35
• You have a point. For such a proof, see here. – Ahaan S. Rungta Nov 7 '13 at 16:36
• It's much simpler here, the sequence is increasing and bounded above by 2. Therefore it converges. The limit does satisfy $x^2 = x+2$, and the reason it's not $-1$ is because the sequence is increasing and starts at a positive number. – Najib Idrissi Nov 7 '13 at 16:39

Set $a_{n+1}=\sqrt{2+a_n}$ and $a_0= \sqrt{2}$. We want to calculate the limit of $a_n$.

First, $0<a_n<2$ implies that $0<a_{n+1} <\sqrt{2+2} =2$ and $0<a_0<2$ then $a_n<2$ for any integer $n\geq 0$.

Second, we have $a_{n+1}^2 -a_n^2= 2+a_n-a_n^2 =(2-a_n)(1+a_n)>0$ then $a_n$ is increasing.

Since $a_n$ is increasing and bounded from above it converges to $a\in[0,2]$ and then $a=\sqrt{2+a}$ and clearly we have $a=2$.

• To rigorously justify the last step of finding the limit, use the fact that subsequence of a sequence converges to the same limit as the original, so $\lim a_{n+1} = \lim a_n = a$; then by the Algebraic Limit Theorem, we can manipulate the limits to simplify: $(\lim (a_{n+1}))^2 = a^2 = \lim (a_{n+1})^2 = \lim (2 + a_n) = 2 + \lim a_n = 2+a$. – Yibo Yang Jul 15 '17 at 5:44