I am looking at an exercise,where given that $a_{n}=\sqrt{2+\sqrt{2+...+\sqrt{2}}}$,I have to show that $\lim_{n \to \infty}a_{n}=2$. We find that $a_{0}=0,a_{1}=\sqrt{2},a_{2}=\sqrt{2+a_{1}} \text{ and so on and we conclude that }a_{n+1}=\sqrt{2+a_{n}}$.Then we take the function $\varphi(x)=\sqrt{2+x}$,in order to apply the Banach fixed-point theorem.But which interval do we have to take?At the solution I have,the interval $[a,b]=[0,3]$ is taken,but I really haven't understood why?Could you explain it to me?
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The interval is taken because it works; there isn't really a deeper reason. One of the nice things about real analysis -- especially when you take limits -- is that it doesn't matter too much what you choose for bounds on things. You usually have a great deal of freedom to make simple choices, or choices that simplify other things.
The argument you want to use only has three requirements:
- It's a closed interval
- Your sequence has a point in the interval
- The function is a contraction mapping on the interval
It's almost hard to find an interval you couldn't use to make this argument work -- especially since you already know that the fixed point you're looking for should be $2$.
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1@evinda: Yep. I confess I would probably choose an upper bound larger than 2, though, just so I don't have to look up the fine print to see if being on the boundary matters. – Jan 14 '14 at 13:27
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I understand..Could you also explain me how you found that $a_{n}<2$ ?Did you suppose it and checked it or did you find it in an other way? – evinda Jan 14 '14 at 13:33
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1@evinda: I actually didn't use that at all, nor does the proof via fixed point theorem that you suggest. It's easy to prove if you already proven the limit to be $2$ since the sequence is increasing -- and if you haven't proven that yet, induction works as the other answers suggest. – Jan 14 '14 at 13:40
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(1) Show inductively that $a_n$ is increasing.
(2) Show inductively that $a_n<2$.
Then, (1)+(2) imply that $a_n$ is convergent. Say $a_n\to x$. But then $a_{n+1}=\sqrt{2+a_n}\to x$. At the same time $\sqrt{2+a_n}\to\sqrt{2+x}$. Hence $x=\sqrt{2+x}$, thus $x^2-x-2=0$. This implies that $x=2$ or $x=-1$. But $x=-1$ is impossible, as $a_n>0$. Thus $a_n\to 2$.
Yiorgos S. Smyrlis
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I edited the value of $a_{1}$.It is equal to $\sqrt{2}$..But,why is it like that: $a_{n}=a_{1}$ ? – evinda Jan 14 '14 at 13:01
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$a_1=\sqrt{2}$, $a_{n+1}=\sqrt{2+a_n}, n=1,2,\cdots$
If we take the sequense $b_1=2$ and $b_{n+1}=\sqrt{2+b_n}, n =1,2, \cdots$
we can see that for every $n$ $a_n \le b_n=2$. Therefore, the sequence $a_n$ is bounded from above. On the other hand $a_1=\sqrt{2} \le a_2=\sqrt{2+\sqrt{2}}$ and by induction we can see that the sequence $a_n$ is increasing. Therefore ite is convergent. If we denote the limit by $a$ and pass to the limit in $a_{n+1}=\sqrt{2+a_n}$ we have that $a=\sqrt{a+2}$. The solution of this equation is 2.
The inequalities above show that $a_n \in (0,3)$ for all $n$.
kmitov
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$a_{n+1}^2=2+a_n\le 2+2=4$ and since all $a_n$ are positive: $a_{n+1}\le{2}$
– Bernd Jan 14 '14 at 13:36