Using BMCT to prove the sequence $a_{n+1} = \sqrt{2 + a_n}$ is bounded and increasing.

Solution 1:

I like the following reasoning. $$a_{n+1}-a_n=\sqrt{2+a_n}-a_n=\frac{2+a_n-a_n^2}{\sqrt{2+a_n}+a_n}=\frac{(2-a_n)(a_n+1)}{\sqrt{2+a_n}+a_n}$$ and after your proof of $a_n<2$ we get that $\{a_n\}$ is increasing.

Solution 2:

Here is a generalization. My original answer is at the end.

If $a_{n+1} =\sqrt{a_n+d^2-d} $ where $d > 1$, then once $a_n < d$ then $a_n \to d$ linearly.

This problem is the case $d = 2$.

If $a_n \lt d$ then $a_{n+1} \lt \sqrt{d+d^2-d} =d $, so that all subsequent $a_n < d$.

Let $a_n = b_n+d$. Want to show that $b_n \to 0$.

Since $0 < a_n < d$, $-d < b_n < 0$ or $d^2-d < b_n+d^2 < d^2$ or $\sqrt{d^2-d} < \sqrt{b_n+d^2} < d $.

Then $b_{n+1}+d =\sqrt{b_n+d+d^2-d} =\sqrt{b_n+d^2} $ so that

$\begin{array}\\ b_{n+1} &=\sqrt{b_n+d^2}-d\\ &=(\sqrt{b_n+d^2}-d)\dfrac{\sqrt{b_n+d^2}+d}{\sqrt{b_n+d^2}+d}\\ &=\dfrac{b_n}{\sqrt{b_n+d^2}+d}\\ \text{so}\\ |b_{n+1}| &=\dfrac{|b_n|}{|\sqrt{b_n+d^2}+d|}\\ &\lt\dfrac{|b_n|}{\sqrt{d^2-d}+d}\\ \end{array} $

so $b_n \to 0$ linearly.

Note that this also shows that $|b_{n+1}| \gt \dfrac{|b_n|}{2d} $, so the convergence is at most linear.

Since $b_n \to 0$, $\dfrac{b_{n+1}}{b_n} \to \dfrac1{2d} $, so this is the exact rate of convergence.

If $a_n \lt 2$ then $a_{n+1} \lt \sqrt{4} = 2$, so the sequence is bounded.

Let $a_n = b_n+2$. Want to show that $b_n \to 0$.

Since $0 < a_n < 2$, $-2 < b_n < 0$ or $2 < b_n+4 < 4$ or $\sqrt{2} < \sqrt{b_n+4} < 2$.

$b_{n+1}+2 =\sqrt{b_n+4} $ or

$\begin{array}\\ b_{n+1} &=\sqrt{b_n+4}-2\\ &=(\sqrt{b_n+4}-2)\dfrac{\sqrt{b_n+4}+2}{\sqrt{b_n+4}+2}\\ &=\dfrac{b_n}{\sqrt{b_n+4}+2}\\ \text{so}\\ |b_{n+1}| &=\dfrac{|b_n|}{|\sqrt{b_n+4}+2|}\\ &\lt\dfrac{|b_n|}{2+\sqrt{2}}\\ \end{array} $

so $b_n \to 0$ linearly.

Note that this also shows that $|b_{n+1}| \gt \dfrac{|b_n|}{4} $, so the convergence is at most linear.

Since $b_n \to 0$, $\dfrac{b_{n+1}}{b_n} \to \dfrac14 $.