Using BMCT to prove the sequence $a_{n+1} = \sqrt{2 + a_n}$ is bounded and increasing. My solution:
Proof of (1)
$$< \sqrt{2 + 2} \text{ by I. H.}$$
$$= 2$$ As required. Therefore by PMI $\{a_n\}$ is bounded above by 2
Proof of (2)
$$(a_n - 2)(a_n + 1) < 0$$
Therefore $\{a_n \}$ is strictly increasing. 
By BMCT it states that if a sequence is bounded above and strictly increasing, then it converges. Therefore this sequence by BMCT converges. 
Is this correct? Is there anything else I should say or add? 
 A: 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.
A: 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
$.
