# 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?

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But how to calculate the exact value of the limit?

Hint. You have shown that the limit does exist, then it has to satisfy
$$l= \sqrt{2 + \sqrt{l}}, \qquad \sqrt{2}<l\le2,$$ by squaring twice one gets
$$l^4-4l^2-l+4=0$$ then, by solving the quartic equation with Ferrari’s method, one gets

$$l=\frac{1}{3\sqrt[3]{2}} \left(79+3 \sqrt{249}\right)^{1/3}+\frac{1}{3\sqrt[3]{2}} \left(79-3 \sqrt{249}\right)^{1/3}-\frac13$$

observing that
$$l= 1.83117720\cdots.$$

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
$$s=-\frac13+\frac{1}{\sqrt[3]{54}} \Big(\left(79+3 \sqrt{249}\right)^{1/3}+ \left(79-3 \sqrt{249}\right)^{1/3}\Big).$$

If $s_1=\sqrt{2}$ and
$$s_{n+1}=\sqrt{2+\sqrt{s_n}}$$
prove 1) that $\{ s_n \}$ converges and 2) that $s_n<2$ for any $n\in \mathbb{N}$..
First we show $\{ s_n\}$ is increasing by induction.

Base case:
$$2>0\Rightarrow 2+\sqrt{2}>\sqrt{2}\Rightarrow \sqrt{2+\sqrt{2}}>\sqrt{2} \Rightarrow s_2>s_1$$
Inductive hypothesis: Suppose $s_n>s_{n-1}$, then
$$\sqrt{s_n}>\sqrt{s_{n-1}}\Rightarrow 2+\sqrt{s_{n}}>2+\sqrt{s_{n-1}}\Rightarrow \sqrt{2+\sqrt{s_{n}}}>\sqrt{2+\sqrt{s_{n-1}}} \Rightarrow s_{n+1}>s_{n}$$
Now, showing part 2 to be true will allow us to conclude part 1, as increasing
sequences bounded above converge.
Base case: $\sqrt{2}<2$.
Inductive hypothesis: Suppose $s_n<2$. Then
$$\sqrt{2+\sqrt{s_{n-1}}}<2\Rightarrow 2+\sqrt{2+\sqrt{s_{n-1}}}<4\Rightarrow \sqrt{2+\sqrt{2+\sqrt{s_{n-1}}}}<2\Rightarrow \sqrt{2+s_n}<2\\ \stackrel{\text{since s_n>1\Rightarrow s_n>\sqrt{s_{n}}>0}}{\Rightarrow}\sqrt{2+ \sqrt{s_n}}<\sqrt{2+s_n}<2\Rightarrow s_{n+1}<2$$
and we can conclude.