Prove monotonicity of a sequence.

Question

$1)$
I have a sequence defined in terms of a recurring relationship. Namely, $$s_0=\sqrt{2},\;s_{n+1}=\sqrt{2+\sqrt{s_n}}.$$ I could prove that it is bounded, but having difficulty in proving that it is monotonic so that I can apply Heine-Borel theorem to prove convergence. Can someone tell me how to do it?

$2)$
Also how to calculate the limit rigorously.
I calculated it using the fact that we can substitute both $s_n$ and $s_{n+1}$ with a supposed limit $l$. While it is intuitive that this approach works, I need a rigorous proof of how such an evaluation works without using concepts of continuity and differentiability as the book (baby Rudin) I am going through didn't reach there yet.

Answer 1

You claime to have proved the boundedness of the sequence.

MONOTONICITY:
Clearly, $s_1>s_0=\sqrt2.$
For $n\geq 1,$ et us compute $$\begin{aligned}s_{n+1}-s_n&=\sqrt{2+\sqrt{s_n}}-\sqrt{2+\sqrt{s_{n-1}}}\\\\&=\frac{\sqrt{s_n}-\sqrt{s_{n-1}}}{\sqrt{2+\sqrt{s_n}}+\sqrt{2+\sqrt{s_{n-1}}}}\\\\&=\frac{s_n-s_{n-1}}{\underbrace{\left(\sqrt{s_n}+\sqrt{s_{n-1}}\right)}_{>0}\underbrace{\left(\sqrt{2+\sqrt{s_n}}+\sqrt{2+\sqrt{s_{n-1}}}\right)}_{>0}}\end{aligned}$$ Edit

Therefore, if $s_{n}-s_{n-1}>0,$ then also $s_{n+1}-s_{n}>0.$ By induction, we conclude that $$\forall n\in\mathbb{N}:\;s_{n}>s_{n-1}.$$ The sequence is increasing.

We have an increasing and bounded sequence, therefore it has a limit.

The way you used to find the limit is not intuitive, it is pretty legitimate:
If the limit exists, then it satisfies the equation $$l=\sqrt{2+\sqrt l}\tag{1}.$$ And we know already, that the limit exists.

The equation $(1)$ rewrites $$l^4-4l^2-l+4=(l-1)(l^3+l^2-3l-4)$$ With the use of Descartes'rule of signs we know that the cubic factor has exactly one positive root. With the use of numerical methods we find $\;l\approx 1.831177>\sqrt 2$

Answer 2

I believe this is Rudin Exercise 3.3? To show that $\{s_n\}$ is monotonic, you can use induction. Indeed, for $n=1$ we have $s_1=\sqrt{2}$ and $s_2=\sqrt{2+\sqrt{\sqrt{2}}}$, so by the monotonicity of the square root function, $s_1<s_2$. Now for any $n\geq 1$, use some algebra to show that $s_n<s_{n+1}$. Then this establishes that $$s_1<s_2<s_3<\cdots$$ and so $\{s_n\}$ is monotonic (in fact strictly increasing). Next, try to show that $s_n<2$ for all $n\geq 1$ (one possible approach is by contradiction: assume that there is a smallest integer $n\geq 2$ such that $s_n\geq 2$, and try to get a contradiction on the minimality of $n$).

At this point you have shown that $\{s_n\}$ is monotonic and bounded. Therefore, by theorem 3.14, $\{s_n\}$ converges. Finding the exact limit can be bit difficult -- it is the greater root of the equation $x^4-4x^2-x+4=0$, but that isn't easy to calculate.

Edit: the induction is as follows

$$s_{n+1}>s_n\iff \sqrt{2+\sqrt{s_n}}>s_n$$ $$\iff2+\sqrt{s_n}>s_n^2=2+\sqrt{s_{n-1}}$$ $$\iff \sqrt{s_n}>\sqrt{s_{n-1}}\impliedby s_n>s_{n-1}.$$

Answer 3

First call $f$ the function $$f:x\mapsto \sqrt{2+\sqrt{x}}\,.$$ $f$ is an increasing function because $g:y\mapsto\sqrt{y}$ and $h:y\mapsto 2+y$ are both increasing and $f = g\circ h\circ g$.

Now solving $f(x) = x$ you can find that there is a unique solution $x_\star$ and that this solution is positive. For $x=0$ you have that $f(0)>0$. So you can infer that $f(x)>x$ for all $x<x_\star$ and $f(x)<x$ for all $x>x_\star$. (To prove this rigorously the easiest way is to use continuity and the fact that the two graphs can cross just once. If you don't want to use this property you can solve explicitly like in ilovebulbasaur's answer.)

For all $s_n<x_\star$ you have $s_{n+1} = f(s_n)>s_n$. By induction you easily show that if $s_0<x_\star$ than all $s_n<x_\star$. This proves that the sequence is monotonic (increasing) if $s_0<x_\star$.

Now if you have find already $x_\star$ you will see that $\sqrt{2}<x_\star$. So the monotonicity is proven.

With an analog argument you prove that the sequence would be decreasing if $s_0<x_\star$.

To prove the existence of the limit you can use that the sequence is bounded and so must have a subsequence which is convergent (Bolzano-Weierstrass). Then since it is monotonic the whole sequence converge. Then it must converge to $x_\star$ because it is the only fixed point.

Answer 4

$x$ is increasing$\implies\sqrt x$ is increasing $\implies2+\sqrt x$ is increasing$\implies\sqrt{2+\sqrt x}$ is increasing.

Now if the sequence converges, it converges to a solution of

$$s=\sqrt{2+\sqrt s}$$ which is a root of

$$(s^2-2)^2-s$$ and exceeds $\sqrt2$.

By inspection $s=1$ is a solution and after long division $$s^3+s^2-3s-4=0$$ remains. Then by the substitution

$$s:=\frac{2\sqrt{10}t-1}3$$ the equation becomes

$$4t^3-3t=\frac{79\sqrt{10}}{200}.$$

Now setting

$$t=\cosh(u),$$

$$\cosh(3u)=\frac{79\sqrt{10}}{200}, \\ t=\cosh\left(\dfrac13\text{arcosh}\left(\dfrac{79\sqrt{10}}{200}\right)\right), \\s=\frac{2\sqrt{10}\cosh\left(\dfrac13\text{arcosh}\left(\dfrac{79\sqrt{10}}{200}\right)\right)-1}3.$$