Limit of $x_n = \sqrt{2+x_{n-1}}$

Question

I am trying to find quite a simple limit i.e. for $x_n = \sqrt{2+x_{n-1}}$ and $x_1 = \sqrt{2}$, find:

$$\lim_{n \to \infty}{x_n}$$

So in order to use recurrence relations to find the limit, I must first prove that this limit exists. So i thought I would prove that this sequence is monotone sequence and bounded from above.

By solving $x_{n+1} > x_n$ I got that this holds for $-1 < x_n <2$ . So i thought immediately that by proving $x_n <2$ proves the existence of this limit.

This part seems to be causing me problems. Proving:

$$\sqrt{2+\sqrt{2+\sqrt{2+...}}} <2$$

Now this is intuitive and very simple but I need a formal proof. I did this: $$\sqrt{2} < 2 $$ Adding 2 to both sides gives: $$2+\sqrt{2} < 4$$ Square rooting : $$\sqrt{2+\sqrt{2}} < 2$$ Repeating the process $n$ times: $$\sqrt{2+\sqrt{2+\sqrt{2}...}} < 2$$

But i am not sure of this proof is formal enough or if it is a valid proof. I believe this problem has been solved before but i couldn't find it on math.stackexchange.com .

Of course, calculating the limit gives: $$\lim_{n \to \infty } {x_n} = 2$$

Summary: Is the proof provided above formal enough, and if there are other proofs, I would like to see them?

You can use induction $x_n < 2 \implies x_{n+1}=\sqrt{x_n+2}<2$ in much the same way... — Macavity, Sep 06 '15 at 07:46
You see this. I think you will get your answer.http://math.stackexchange.com/questions/1418807/proving-convergence-of-a-series-and-then-finding-limit/1418830#1418830 — , Sep 06 '15 at 07:46
@SubhadeepDey Yes, thank you. I know this is a common problem so i knew there already is a solution here , but i just couldn't find it :D — Transcendental, Sep 06 '15 at 07:50
@Macavity Well, yes, I basically did use the induction there. Thanks :D — Transcendental, Sep 06 '15 at 07:51

score 0 · Accepted Answer · answered Sep 06 '15 at 08:01

0

Here is another proof: $0<|x_n-2| = \dfrac{|x_{n-1}-2|}{\sqrt{x_{n-1}+2}+2}<\dfrac{|x_{n-1}-2|}{3}<\dfrac{|x_{n-2}-2|}{9}<\cdots <\dfrac{|x_1-2|}{3^{n-1}}<\dfrac{1}{3^{n-1}}$, and apply squeeze theorem, the limit is $2$ as claimed.

answered Sep 06 '15 at 08:01

DeepSea

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Limit of $x_n = \sqrt{2+x_{n-1}}$

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