Recursive sequence $a_0= 1$ and $a_{n+1}=\sqrt{2+a_n}$

Question

I have a question and hope for some help, because I not so sure somehow..

QUESTION:

We have a sequence defined by $a_0= 1$ and $a_{n+1}=\sqrt{2+a_n}$ for $n\in \mathbb{N}$. Show with induction that, for all $n\in \mathbb{N}$, $0\leq a_n\leq 2$ and $a_n\leq a_{n+1}$. Decide if the sequence is convergent and show its limit.

SOLUTION:

So, at first I start with induction, do I need to start with $0$ or $1$? of course both if them pass for it. by $n=0$ we have $0\leq a_0=1 \leq 2$ and for $n=1$ we have $0\leq a_2=\sqrt{3} \leq 2$ .

Then for $n+1$ we have $0\leq a_n \leq 2\implies2\leq2+a_n\leq4\implies\sqrt2\leq\sqrt{2+a_n}\leq2\implies\sqrt2\leq a_{n+1}\leq2)$ right? So the first part is done, I guess.

Next part is $a_{n+1}-a_n=\sqrt{2+a_n}-a_n=\frac{(\sqrt{2+a_n}-a_n){(\sqrt{2+a_n}+a_n)}}{{(\sqrt{2+a_n}+a_n)}}=\frac{\left(\sqrt{2+a_n}\right)^2-a_n^2}{\sqrt{2+a_n}+a_n}$

and ${a_{n+1}-a_n}=\frac{(2+a_n)-a_n^2}{\sqrt{2+a_n}+a_n}=\frac{(1+a_n)(2-a_n)}{\sqrt{2+a_n}+a_n}>0$ everything is positive because of $\sqrt2\leq a_n\leq2$ and $a_{n+1}-a_n>0 \implies a_{n+1}>a_n$.

And the last part would be convergence: Our sequence is monotonic increasing and bounded so there is a limit. $a=\sqrt{2+a}\implies a^2=2+a\implies a^2-a-2=0\implies (a-2)(a+1)=0\implies a=2$ Another root $a=-1$ is wrong because for all $a_n\ge\sqrt2$.

Is my solution okay? Thanks in advance : )

Answer 1

$a_0 = 1, a_1 = \sqrt{3}.$

So, $~a_1 > 0.$

Inductively assume that $~a_n > 0.$
This implies that $~(2 + a_n) > 0 \implies a_{n+1} = \sqrt{2 + a_n} > 0.$

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$~a_1 < 2.$

Inductively assume that $~a_n < 2.$
This implies that $~(2 + a_n) < 4 \implies a_{n+1} = \sqrt{2 + a_n} < 2.$

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$a_1 > a_0 > 0.$

Inductively assume that $a_{n+1} > a_n > 0.$
This implies that $~(2 + a_{n+1}) > (2 + a_n) > 0 \implies $
$\sqrt{2 + a_{n+1}} > \sqrt{2 + {a_n}} > 0 \implies $ $a_{n+2} > a_{n+1} > 0.$

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Therefore, the sequence is strictly positive, strictly increasing, and bounded above by $~2.$

Therefore, the sequence must be convergent.

Let $~L~$ denote the limit of this sequence.

So, $a_n$ goes to $~L,~$ as does $~a_{n+1} = \sqrt{2 + a_n}.$

Therefore,

$$L = \sqrt{2 + L} \implies L^2 = L + 2 \implies (L-2)(L+1) = 0.$$

Since the limit, $~L,~$ must be positive, you must have that $~L = 2.$

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Addendum

My answer begs the following question.

Given that the sequence must be convergent, with some unique limit $~L,~$ and given that this implies that $~a_n \to L,~$ and $~a_{n+1} = \sqrt{2 + a_n} \to L,~$ can you then assert that $~L = \sqrt{2 + L}?$

In my opinion, it depends on whether the instructor/class has presented on point worked examples or a Real Analysis theorem, that justifies this step.

Presumably, you know that the function $~f(x) = \sqrt{x}~$ is a continuous function, and that the composition of two continuous functions is a continuous function. Therefore, you know that the function $~g(x) = \sqrt{2 + x}~$ is a continous function.

This implies that as $~x \to L,~$ that $~g(x) \to g(L).~$

So, as $~a_n~$ goes to $~L,~ g(a_n)~$ goes to $~g(L).$ Further, $~g(a_n) = a_{n+1},~$ which (also) goes to $~L.~$

So, you can conclude that

• $~g(a_n) = a_{n+1}~$ goes to $~L.$

• $~g(a_n) = \sqrt{2 + a_n}~$ goes to $~g(L) = \sqrt{2 + L}.$

Therefore, $~L = g(L) = \sqrt{2 + L}.$

Answer 2

Prove: $a_0=1. 0\le a_n\le 2. \ a_n\le a_{n+1}$ when $a_{n+1}=\sqrt{2+a_n}$.

$\sqrt{2+1}=1.7>1.0$ check!

$\sqrt{2+\sqrt{2+u}}>\sqrt{2+u}$?

$2+\sqrt{2+u}>2+u \iff \sqrt{2+u}>u$

$a_{n+1}>a_n. $QED-I.

Let $u=2-\epsilon, 0<\epsilon<1$.

$ \sqrt{2+u}=\sqrt{4-\epsilon}\le 2. $ QED-II

Monotonic and bounded implies convergence.

Suppose $f(x)=x$ and $|f'(x)|<1$ $|f(x+\delta)-f(x)|=|f(x+\delta)-x|<\delta |f'(x)|<\delta$

So $f(x+\delta)$ is closer to $x$ than $x+\delta$ was.

$f(x)=\sqrt{2+x}\implies f'(x)=\frac{1}{2\sqrt{2+x}}<1$ for $0<x<2$.

This suggests the limit is a positive root of $x=\sqrt{2+x}$. So $x =2$.