Show that $\sqrt{2+\sqrt{2+\sqrt{2...}}}$ converges to 2

Question

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?

http://math.stackexchange.com/questions/849274/the-convergence-of-sqrt-2-sqrt-2-sqrt-2-ldots — 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

Ahaan S. Rungta · Accepted Answer · 2014-08-24T16:00:37.483

17

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 $2$.

edited Aug 24 '14 at 16:00

answered Nov 07 '13 at 16:29

Ahaan S. Rungta

7,706

26

You haven't proven convergence. – Najib Idrissi Nov 07 '13 at 16:33
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If it didn't converge, I wouldn't get a valid solution. – Ahaan S. Rungta Nov 07 '13 at 16:34
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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 07 '13 at 16:35
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You have a point. For such a proof, see here. – Ahaan S. Rungta Nov 07 '13 at 16:36
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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 07 '13 at 16:39
Thanks Ahaan. But I still cannot see that the limit x is $\sqrt{x+2}$. – Yihan Nov 07 '13 at 16:40
Write before you posted, I was about to add this link, which is slightly more appropriate. Your reason about the radical not being $-1$ is identical to mine. – Ahaan S. Rungta Nov 07 '13 at 16:40
Is there a way to show that 2 is the greatest upper bound and then use the monotone convergence theorem? – Yihan Nov 07 '13 at 16:42
@Yihan Note that $x^2-2$ is the same nested radical, since it's infinite. – Ahaan S. Rungta Nov 07 '13 at 16:43
Just curious, why was this downvoted? – Ahaan S. Rungta Nov 09 '13 at 00:23
@AhaanRungta Did I? I might press the wrong button. Sorry – Yihan Nov 09 '13 at 22:55
Your answer is obvious, but I don't think it answers the question. – JobHunter69 Jan 22 '18 at 17:56

score 9 · Answer 2 · answered Aug 24 '14 at 16:18

9

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 $a=2$.

answered Aug 24 '14 at 16:18

Tulip

4,876

1

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 05:44

Show that $\sqrt{2+\sqrt{2+\sqrt{2...}}}$ converges to 2

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