What is the limit of this sequence? Problem 3 in the Exercises after Chapter 3 in Principles of Mathematical Analysis by Walter Rudin, 3rd edition: 
Let $s_1 \colon= \sqrt{2}$, and let 
$$s_{n+1} \colon= \sqrt{2+\sqrt{s_n}} \mbox{ for } n = 1, 2, 3, \ldots. $$ 
Then how to rigorously calculate $$\lim_{n\to\infty} s_n,$$ justifying each step from the definition of convergence of a sequence and the theorems on convergence of sequences as have been proved by Rudin upto this point in his book?  
I know that this sequence is increasing and bounded (above) and hence convergent. 

As you have mentioned already, $L = \lim_{n \rightarrow \infty} s_n$ exists in $\mathbb{R}$. Also $L \geq 0$. Then
$$ L = \sqrt{2 + \sqrt{L}} $$
$$ L^2 = 2 + \sqrt{L} $$
Let $k = \sqrt{L}$. We have
$$k^4 = 2 + k$$
$$(k + 1)(k^3 - k^2 + k - 2) = 0$$
Since $k \geq 0$, we have
$$ k^3 - k^2 + k - 2 = 0$$
which has one positive real root around 1.35. Then $L = k^2$ is around 1.83.
For an exact value of $L$, please see Claude Leibovici's answer.
