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I know this question has been asked many times, but I need a specific part of it.

When we get to the limit part, I had it written like this: $\lim_{n \to +\infty} a_{n+1} = \sqrt{2+\lim_{n \to +\infty} a_n}$

What is the reason I am allowed to put the limit under the square root? Why am I just allowed to put it there? I know it's true, but I am not sure why. I need to justify that in order to receive credit. My instructor said it is a "special word." I have no idea what the word is.

Arctic Char
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Nolan P
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    Continuity. But you need to show that the limit exists first. – copper.hat Sep 11 '20 at 20:21
  • @Sil: Though relevant, the OP is more nuanced than the simpler statement in the link you claim the OP is a duplicate of. The OP is more about fixed points of a given function. – Mittens Sep 11 '20 at 21:49
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    @OliverDiaz please read the question again! It clearly asks about "What is the reason I am allowed to put the limit under the square root" and starts with "I know this question has been asked many times"! Otherwise, it would have been closed as a duplicate of this question. – rtybase Sep 11 '20 at 22:29
  • @rtybase: I READ the question. Now if believe an question is a duplicate (in this case, it is) and the question should be closed, then at least the correct duplicate should be referenced. Sil's request points to a link that is related but does not fully addresses the OP. The link you provide is more appropriate. – Mittens Sep 11 '20 at 23:26
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    @OliverDiaz so you do agree that the question is about "What is the reason I am allowed to put the limit under the square root?" And "continuity" is the "special word"? "Limit exists" contains 2 words :) – rtybase Sep 11 '20 at 23:32
  • @rtybase: of course! but the continuity of $f(x)=\sqrt{2+x}$ is only part of the reason in this instance. The main justification being that the limit actually exists. – Mittens Sep 11 '20 at 23:36

2 Answers2

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IF the limit exists, it will be a fixed point of the function $\sqrt{2+x}$, in other words a solution to the equation $$x=\sqrt{2+x}$$ Here's why: $$\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\sqrt{2+a_n}$$ Given two functions $f$ and $g$, as long as $f$ is continuous and $\lim_{x\to x_0}g(x)$ exists, then $$\lim_{x\to x_0}f(g(x))=f\left(\lim_{x\to x_0}g(x)\right)$$ This can be shown fairly routinely with the $\epsilon ,\delta$ definition of the limit.

Since $\sqrt{2+x}$ is continuous on its domain, $$\lim_{n\to\infty}\sqrt{2+a_n}=\sqrt{2+\lim_{n\to\infty}a_n}=\lim_{n\to\infty}a_{n+1}$$ Since $x:=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}a_n$, the initial statement follows.

Mittens
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K.defaoite
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  • The interesting part is to show that the limit exists! – Mittens Sep 11 '20 at 22:19
  • Your argument does not solve the important question of existence of the limit. – Mittens Sep 11 '20 at 22:21
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    @OliverDiaz The OP did not ask about the existence of the limit.... – K.defaoite Sep 11 '20 at 22:42
  • @K.defaoite: He asked for a justification. The mere continuity of $x\mapsto\sqrt{x+2}$ is not enough. Of course. if the limit exists (as you assume) things are honky dory, but the main part of the justification for $\lim_n a_{n+1}=\sqrt{2+\lim_na_n}$ here is existence. Otherwise there fixed point theorems would be a trivial thing. – Mittens Sep 11 '20 at 22:52
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We have a few claims:

Claim: The sequence $\{ a_{n}\}_{n = 1}^\infty$ is bounded above by $2$.

Proof. The proof can be done by induction. I'll just write the induction step $k + 1$. $$ a_{k + 2}^2 = 2 + a_{k + 1} < 2 + 2 \implies a_{k + 2} < 2 $$ The rest is as goes with the formal writing.

Lemma: $\{ a_n\} = \left\{ 2 \cos \frac{\pi}{2^{n + 1}} \right\} $

Proof. Similar by induction. Assume $$ a_k = 2\cos \left( \frac{\pi}{2^{k + 1}} \right) $$ Then $$ a_{k +1}^2 = 2 + 2 \cos\left( \frac{\pi}{2^{k + 1}} \right) = 2^2 \cos^2 \left( \frac{\pi}{2^{k + 2}} \right) $$ $$ \implies a_{k + 1} = 2 \cos \left( \frac{\pi}{2^{k + 2}} \right) $$ Hence inductively we are done.

Now, we have a specific closed form. Then we have that $\lim\limits_{n \to \infty} a_n$ exists, as $\cos \left( \dfrac{\pi}{2^{n + 1}} \right) \to \cos 0 = 1 $, as $\cos$ is a continuous function on $\left[0, \dfrac{\pi}{2} \right]$.

Thus finally we have that $$ \lim_{n \to \infty} a_n = \lim_{n \to \infty} 2\cos \left( \frac{\pi}{2^{n + 1}} \right) = 2 $$

Edit: I have tried to find a closed-form.

Ralph Clausen
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