How to show that this sequence converges to this limit

Question

The recursive sequence given by

$ s_1 = \sqrt{2} $

$ s_{n+1} = \sqrt{2 + s_n} $

seems to converge to $ 2 $

It's easy to prove that this sequence indeed converges and that $ 2 $ is a upper bound for that sequence, but how to prove that it really does converge to 2?

thanks in advance.

Answer 1

Let $r_{n+1} = \sqrt{4-2^{1-n}}$. Then, $r_1 = \sqrt{2} \leq \sqrt{2} = s_1$.

Further, assume that $r_{n+1} = \sqrt{4-2^{1-n}} \leq \sqrt{2+s_n} = s_{n+1}$.

Then, $s_{n+2} = \sqrt{2+s_{n+1}} = \sqrt{2+\sqrt{2+s_n}} \geq \sqrt{2+\sqrt{4-2^{1-n}}}$ by hypothesis.

Since $n \geq 0 \geq -1$,

$1-n \geq -2n \Rightarrow 2^{1-n}\geq 2^{-2n} \Rightarrow 4+2^{2-n}-2^{1-n} \geq 4+(2^{-n})^2 \Rightarrow 4-2^{1-n}\geq (2-2^{-n})^2$.

Thus, $\sqrt{2+\sqrt{4-2^{1-n}}} \geq \sqrt{2+2-2^{-n}} = \sqrt{4-2^{-n}} = r_{n+2}$.

Thus, for all $n \geq 0$, $r_{n+1} \leq s_{n+1}$.

By squeeze theorem, since $r_{n+1} \rightarrow 2$ and $s_{n+1} \rightarrow L \leq 2$, $s_{n+1}\rightarrow 2$.

Answer 2

You know that the sequence has a limit $L$

Let $ n\to \infty$ and apply the result to $$s_{n+1} = \sqrt{2 + s_n}$$

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

Upon solving for $L$, we get two solutions. $$ L= 2$$ and $$ L=-1$$ $L=-1$ is out of question because all our terms are positive.

Thus $$L=2$$

Answer 3

Suppose that you do not know that the square root is continuous (or even the sequential characterization of continuity). Then we can do the following to show that if it converges to $2$:

We can rewrite the inductive step as $s_{n+1}^2 = 2 + s_n$. Since these two sequences are the same, their limits will be the same. Suppose $s_n \to L$. Then taking the limit of both sides of the equation, we get that $L^2 = 2 + L$. Solving, we get that either $L=-1$ or $L = 2$. Since the sequence is always positive, $L = 2$.