Finding the limit of the recursive sequence $r_{n+1} = \sqrt{2 + r_n}$

Question

In the example I am given, I am told that $r_n$ is defined as:

$$ r_n = \begin{cases} r_0 = \sqrt{2} \\ r_{n + 1} = \sqrt{2 + r_n} \\ \end{cases} $$

I was told to calculate $r_3$ and I found that to be:

$$ r_3 = \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2}}}} $$

Then I am told to find

$$ \lim\limits_{n \rightarrow \infty} r_n $$

So if I let that equal L I get

$$ L = \sqrt{2 + L} $$

This is where I get stuck. I know that the limit is 2 as I have added a 10 terms together but that will not help me in my exam. How should I go about proving this?

Answer 1

First, you need to solve $L=\sqrt {2+L}$ Square it and you have a quadratic. Once you get the solutions, plug them into the original equation to see which one is not spurious. You also have to show that the limit exists. In this case, you can show that (if $r_n \lt 2$, then $r_{n+1} \lt 2$) and $r_{n+1} \gt r_n$ so you have a monotonic sequence bounded above. That gives you convergence.

Answer 2

So... you're asking how to solve the equation $L = \sqrt{2+L}$?

Answer 3

Let $f(x) = \sqrt{2+x}$. Note that if $x \ge \sqrt{2}$, then $f(x) \ge \sqrt{2}$, so $f:[\sqrt{2},\infty) \to [\sqrt{2},\infty)$.

Furthermore, $|f'(x)| \le \frac{1}{2}$ for $x \ge \sqrt{2}$. Hence $f$ is a contraction and the sequence $x_{n+1} = f(x_n)$ will converge to the unique fixed point (in $[\sqrt{2},\infty)$).

To find the fixed point we solve $f(x) = x$. Squaring gives $f(x)^2 = x+2 = x^2$, and the only positive solution of this quadratic is $x=2$.