Prob. 3, Chap. 3 in Baby Rudin: If $s_1 = \sqrt{2}$, and $s_{n+1} = \sqrt{2 + \sqrt{s_n}}$, what is the limit of this sequence? Here's Prob. 3, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition: 

If $s_1 = \sqrt{2}$, and $$s_{n+1} = \sqrt{2 + \sqrt{s_n}} \ \ (n = 1, 2, 3, \ldots),$$ prove that $\left\{ s_n \right\}$ converges, and that $s_n < 2$ for $n = 1, 2, 3, \ldots$. 

My effort: 
We can show that $\sqrt{2} \leq s_n \leq 2$ for all $n = 1, 2, 3, \ldots$. [Am I right?]
Then we can also show that $s_n < s_{n+1}$ for all $n = 1, 2, 3, \ldots$. [Am I right?]
But how to calculate the exact value of the limit? Where does this sequence occur in applications? 

If the limit $s$ of $s_n$ exists, then
$$
s=\lim s_n=\lim s_{n+1}=\lim \sqrt{2+\sqrt{s_n}}=\lim \sqrt{2+\sqrt{s}}.
$$
Hence, the limit satisfies the equation
$$
s=\sqrt{2+\sqrt{s}}.
$$
Thus,
$$
s^2=2+\sqrt{s}\qquad\text{or equivalently}\qquad (s^2-2)^2=s.
$$
Thus the limit is the unique solution of $s^4-4s^2-s+4=0$ which lies in the interval $[\sqrt{2},2]$.
To solve this equation, first observe that $s=1$ is a solution, and hence
$$
s^4-4s^2-s+4=(s-1)(s^3+s^2-3s-4).
$$
We can use this method to solve $\,s^3+s^2-3s-4=0,\,$ and obtain that
$$
s=-\frac13+\frac{1}{\sqrt[3]{54}} \Big(\left(79+3 \sqrt{249}\right)^{1/3}+ \left(79-3 \sqrt{249}\right)^{1/3}\Big).
$$
