Proof $\lim\limits_{n \rightarrow \infty} {\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}=2$ using Banach's Fixed Point

Question

I'd like to prove $\lim\limits_{n \rightarrow \infty} \underbrace{\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}_{n\textrm{ square roots}}=2$ using Banach's Fixed Point theorem.

I think I should use the function $f(x)=\sqrt{2+x}$. This way, if I start the iterations for example with $x_0=0$, I will have $x_1=\sqrt2$. When I calculate $x_2$ I will get $\sqrt{2+\sqrt{2}}$. And $x_3 = \sqrt{2+\sqrt{2+\sqrt{2}}}$ and so on. I can see that these iterations are monotone increasing, but how can I show that this converges to 2?

Pseudo-related formula I found: http://en.wikipedia.org/wiki/Vieta_formula

Many thanks in advance!

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Following clark's advice, here's my proof this is a contraction. I'm using the interval $D=[0, 2]$.

$f'(x)=\frac{1}{2\sqrt{x+2}}$, which is monotone decreasing. This means its highest value in $D$ is $0$. $f'(0)=\frac{1}{2\sqrt{2}} < 1$. The rate $M$ of the contraction is then $\frac{1}{2\sqrt{2}}$.

Answer 1

In order to use Banach's fixed pointed theorem you have to show $ |f(x)-f(y)| < M(|x-y|)$ in some interval, say $[a,b]$. Then your work would to prove that starting with $x_0=c\,\,$ then$x_{n+1}=f(x_n)$ stays in that interval, i.e.: $a \leq x_n \leq b$,(so your function is well defined $f:[a,b] \rightarrow \mathbb{R}$. That $M $ can be found $ f'(y_0) = M$ and bound the derivative. Then you will know the limit is the solution $f(k)=k$

EDIT: Since you took the interval [0,2] you need to prove that for $y \in [0,2] $ $0 \leq f(y) \leq 2$ the first holds trivially. For the second you have $\sqrt{2+ \sqrt {2}} \leq 2 \Leftrightarrow \sqrt {2} \leq 2$ which holds. Now you are done because you have that every $x_n$ stays in the interval you choosed. So Banach's fixed point theorem can be applied. (Note that you defined $f$ on $D$ so the previous step is to make sure that the $f$ you took is well defined, because every $x_n$ is used by $f$ to define $x_{n+1}$).

Answer 2

This doesn't answer your question, but it might be of interest:

Define que sequence $\{x_n\}$ by

$$\begin{cases} x_0=0 \cr x_n = \sqrt{k+x_{n-1}}\end{cases}$$

with $k>0$

I claim that $$\lim_{n \to \infty}x_n=r$$

where $r$ is the positive root of the equation

$$\tag A x^2-x-k=0 $$

PROOF

$(1)$ The sequenece is increasing. By induction:

It is true for $x_0=0,x_1=\sqrt k$. Assume true for $k=1,2,\dots,n$, then

$$x_n > x_{n-1} \Rightarrow x_n+k > x_{n-1}+k \Rightarrow$$

$$\Rightarrow \sqrt{x_n+k} > \sqrt{x_{n-1}+k} \Rightarrow x_{n+1} > x_n$$

$(2)$ The sequence is bounded above by $r$. By induction:

It is true for $n=0,1$. Assume true for $k=1,2,\dots,n$, then

$$x_{n} < r$$

$$x_{n}+k < r+k$$

$$\sqrt{x_{n}+k} < \sqrt{r+k}=r$$

since $r$ satisfies $(A)$.

Then by the Monotone Convergence Theorem, the sequence has a limit. In particular, this means that $\ell = \lim x_n = \lim x_{n-1}$, so that

$$\lim_{n \to \infty} x_n = \lim_{n \to \infty}\sqrt{x_{n-1}+k} $$

$$\lim_{n \to \infty} x_n = \sqrt{ \lim_{n \to \infty} x_{n-1}+k} $$

$$\ell = \sqrt{\ell+k} $$

$$\ell^2-\ell -k = 0 $$

Then either

$$\ell_1 = \frac{1+\sqrt{1+4k}}{2}$$

or

$$\ell_2 = \frac{1-\sqrt{1+4k}}{2}$$

But the latter is impossible since $\ell_2 <0$. It follows that

$$\ell_1 = r$$ the positive root of the equation $x^2-x-k=0$. $\blacktriangle$

Your problem is the special case $k=2$, which yields

$$\ell = \frac{1+\sqrt{1+4\cdot 8 }}{2}=2$$