Limit of $\{a_n\}$, where $a_{n+1} = \sqrt{2+a_n}$

Question

I am struggling with this question:

Let $\{a_n\}$ be defined recursively by $a_1=\sqrt2$, $a_{n+1}=\sqrt{2+a_n}$. Find $\lim\limits_{n\to\infty}a_n$. HINT: Let $L=\lim\limits_{n\to\infty}a_n$. Note that $\lim\limits_{n\to\infty}a_{n+1}=\lim\limits_{n\to\infty}a_n$, so $\lim\limits_{n\to\infty}\sqrt{2+a_n}=L$. Using the properties of limits, solve for $L$.

I just don't know how I am suppose to find the limit of that or what my first step is. Any help?

Note that $a_{n}$ is increasing and bounded by $2$ so it tends to a limit and on solving $\sqrt{2 + L} = L$ we get $L = 2$. — Paramanand Singh, Jul 01 '15 at 03:49
Hint for a different approach: Prove by induction that $a_n=2\cos(\pi/2^{n+1})$ and use the continuity of cosine. — Jyrki Lahtonen, Jul 01 '15 at 06:14
@Paramanand Singh: I added a detailed proof of your statement as another answer. — marty cohen, Jul 01 '15 at 22:00
You can find several closely related posts on this site, for example: Show $a_{n+1}=\sqrt{2+a_n},a_1=\sqrt2$ is monotone increasing and bounded by $2$; what is the limit?, $\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$ and Proof Using the Monotone Convergence Theorem for the sequence $a_{n+1} = \sqrt{4 + a_n}$ — Martin Sleziak, Jul 01 '15 at 23:09
possible duplicate of $\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$ — Jonas Meyer, Jul 03 '15 at 05:52

score 3 · Accepted Answer · answered Jul 01 '15 at 03:55

3

If you have a recursion of the form $a_{n+1} =f(a_n) $, if $L = \lim_{n \to \infty} a_n $, then we must have $L = f(L)$.

In your case, $f(x) = \sqrt{2+x}$.

Therefore, for any limit $L$, we must have $L = \sqrt{2+L}$.

Squaring, $L^2 = L+2$, which is a standard quadratic equation.

Completing the square, from $L^2-L = 2$ we get $L^2-L+1/4 = 2+1/4 =9/4$, so $(L-1/2)^2 = 9/4$.

By a miracle of homework problems, the right side is a square, so $(L-1/2)^2 = 9/4 = (3/2)^2$.

Taking square roots, and remembering that square roots can be negative as well as positive, we get $L-1/2 = \pm 3/2$. Therefore $L = 1/2+3/2 = 2$ or $L = 1/2 - 3/2 = -1$.

Since $L$ must be positive, we reject the negative solution, which goes into the corner and pouts.

This leaves only $L=2$, and we see by substitution that this does satisfy $L = \sqrt{2+L}$.

And they all lived happily ever after, at least until the next problem.

answered Jul 01 '15 at 03:55

marty cohen

110,450

Thank you! So this is a different problem I had: http://www.tiikoni.com/tis/view/?id=34f18a0
I am getting really strange looking 1st 4 numbers. Is that it?
– Saturn Indra Jul 01 '15 at 04:32
$a_1 = \sqrt{2}$, $a_2 = \sqrt{2+\sqrt{2}}$, $a_3 = \sqrt{2+ \sqrt{2+\sqrt{2}}}$, $a_4 = \sqrt{2+ \sqrt{2+ \sqrt{2+\sqrt{2}}}}$. – marty cohen Jul 01 '15 at 04:36
Thank you! So I wasn't wrong after all. It just looked really unusual to me. – Saturn Indra Jul 01 '15 at 04:37
Nested square roots have some interesting and surprising properties. Do a search. Here is one result: https://en.wikipedia.org/wiki/Nested_radical – marty cohen Jul 01 '15 at 04:39
2

Are you sure that your first sentence holds without the assumption that $f$ is continuous at $L$? Anyway, you (as well as the other answerers and the hint) seem to ignore the question whether the limit exists :-) – Jyrki Lahtonen Jul 01 '15 at 06:12
Not so. I said that if the limit exists, then it must satisfy $L = f(L)$. Once $L$ has been found, the next step is to prove that it is the limit. Of course, for some problems, $L$ can not be explicitly found. The usual method then is to prove that the sequence is monotonic and bounded. That can be done in this case. – marty cohen Jul 01 '15 at 21:49

Chiranjeev_Kumar · Answer 2 · 2015-07-01T04:03:33.943

1

$$L=\sqrt{2+L}$$, squaring both side we have,

$$L^2-L-2=0$$

$$(L-2)(L+1)=0$$

which gives $L=2,-1$

since $a_n\gt 0$, for all $n\in \Bbb N$, Hence $L=2$

edited Jul 01 '15 at 04:03

answered Jul 01 '15 at 03:53

Chiranjeev_Kumar

3,079
17
32

score 0 · Answer 3 · answered Jul 01 '15 at 21:58

This expands a comment of mine to prove that the sequence is monotonic and bounded and, therefore, converges.

Suppose $a < 2$. Then I want to show that $a < \sqrt{2+a} < 2$.

Let $2-a = d > 0$. Then $a = 2-d$. $\sqrt{2+a} = \sqrt{2+2-d} = \sqrt{4-d} < 2 $. Also, since $(1-z)^2 = 1-2z + z^2 > 1-2z $, $\sqrt{1-2z} > 1-z $. Therefore $\sqrt{2+a} =\sqrt{4-d} = 2\sqrt{1-d/4} > 2(1-d/2) =2-d =a $.

Since $a_1 = \sqrt{2} < 2$, all following $a_n$ satisfy $a_n < 2$ and $a_n < a_{n+1} $.

Limit of $\{a_n\}$, where $a_{n+1} = \sqrt{2+a_n}$

3 Answers3