Can any one tell me how to prove that: The sequence $x_{1}=\sqrt{2}$ , $x_{2}=\sqrt{2+\sqrt{2}}$, $\cdots$, $x_{n}=\underbrace{\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}_{n \text{ times}}$ converges and compute its limit.
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Below are the steps you need to work out.
First note that we have $$x_{n+1} = \sqrt{2+x_n}$$ with $x_0 = \sqrt2$.
$1$. Prove that $x_n \in [\sqrt2,2)$ by induction.
$2$. Prove that $x_n$ is an increasing sequence by induction.
$3$. Now let $\lim_{n \to \infty} x_n = L$ and use properties of limits to conclude that $$L = \sqrt{2+L}$$
$4$. Use the fact that $L > 0$ to conclude that $L=2$.
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Let denote by $$f(x)=\sqrt{2+x}$$ then we can verify easily that $x^*=2$ is the positive fixed point for $f$ i.e $f(x^*)=x^*$.
Now $f$ is increasing function and since $x_0\leq x_1\leq x^*$ then $x_1=f(x_0)\leq x_2=f(x_1)\leq x^*=f(x^*)$ and by a simple induction we find $$x_{n}=f(x_{n-1})\leq x_{n+1}=f(x_n)\leq x^*=f(x^*)$$ hence the sequence $(x_n)$ is increasing and bounded above by $x^*$ so it's a convergent sequence and its limit is the fixed point $x^*$.