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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?

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closed as off-topic by graydad, Claude Leibovici, Ali Caglayan, user91500, Jonas Meyer Jul 3 '15 at 5:52

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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.

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  • $\begingroup$ Thank you! So this is a different problem I had: tiikoni.com/tis/view/?id=34f18a0 I am getting really strange looking 1st 4 numbers. Is that it? $\endgroup$ – Saturn Indra Jul 1 '15 at 4:32
  • $\begingroup$ $a_1 = \sqrt{2}$, $a_2 = \sqrt{2+\sqrt{2}}$, $a_3 = \sqrt{2+ \sqrt{2+\sqrt{2}}}$, $a_4 = \sqrt{2+ \sqrt{2+ \sqrt{2+\sqrt{2}}}}$. $\endgroup$ – marty cohen Jul 1 '15 at 4:36
  • $\begingroup$ Thank you! So I wasn't wrong after all. It just looked really unusual to me. $\endgroup$ – Saturn Indra Jul 1 '15 at 4:37
  • $\begingroup$ Nested square roots have some interesting and surprising properties. Do a search. Here is one result: en.wikipedia.org/wiki/Nested_radical $\endgroup$ – marty cohen Jul 1 '15 at 4:39
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    $\begingroup$ 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 :-) $\endgroup$ – Jyrki Lahtonen Jul 1 '15 at 6:12
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$$L=\sqrt{2+L}$$, 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 $L=2$

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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} $.

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