Evaluating $\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \cdots }}}}$

Question

I saw this problem in a math competition and I have no idea how to solve it:

$$\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \cdots }}}}$$

Answer 1

$$x =\sqrt{2+\sqrt{2+\sqrt{2}}+...}$$ $$x =\sqrt{2+x}$$ $$x^2= 2 + x$$ You just need to solve quadratic equation and get $x=2$.

Answer 2

We need to understand what we mean by this expression formally. We can formalize it in a following way. Consider a sequence: $$x_n = \sqrt{2+x_{n-1}},\,x_1 = \sqrt{2}$$ Then what we want to find, is $$\lim_{n\to\infty} x_n.$$ We can prove that this sequence is increasing. $$x_n-x_{n-1} = \frac{2+x_{n-1}-x_{n-1}^2}{\sqrt{2+x_{n-1}}+x_{n-1}} $$ The denominator is positive, we need to take care only about the nominator. $$2+x_{n-1}-x_{n-1}^2 = (1+x_{n-1})(2-x_{n-1})>0 $$ This is because $x_n<2$ which can be proved by induction. The sequence is bounded and increasing, thus we know that it must converge. It must also be a root of $$x = \sqrt{2+x}$$ which can be seen by taking limit of both sides in equation $x_n = \sqrt{2+x_{n-1}}$. This leaves us with $x=2$, so we must have $\lim_{n\to\infty} x_n = 2$.

Answer 3

This is not an answer, but a graphical comment (so please don't downvote).

Here's a graph of the value of the expression as you include more and more of the terms (square roots):

Indeed, it asymptotes at $2$, as shown above.