Suppose $a_0$ is an arbitrary positive real number. Define the sequence $\{a_n\}$ by $$a_{n+1}=\frac{a_n+2}{a_n+1}$$ for all $n\geq0$. I have to prove that $\{a_n\}$ converges.
My attempt: If $a=\lim_{n\to\infty}{a_n}$ exists, then it should be a solution to $$a=\frac{a+2}{a+1}$$ which is $\sqrt2$. Thus I need to show that $|\sqrt2 - a_n|$ gets arbitrarily small for large $n$. I tried to prove that $|\sqrt2-a_n|<|\sqrt2-a_{n-1}|$ but couldn't.
Your approach works, with some care: we see that: \begin{align*} |a_{n+1}-\sqrt{2}|&=|\frac{a_n(1-\sqrt{2})+(2-\sqrt{2})}{a_n+1}| \\ &\le |\sqrt{2}-1||a_n-\sqrt{2}| \end{align*}Letting $|a_0-\sqrt{2}|=c$, this proves that $|a_n-\sqrt{2}|\le c|\sqrt{2}-1|^n$, which goes to $0$.
This sequence is defined with a decreasing function $f$ and has positive terms. Hence, if $a_{n-1}<a_n$, $\;a_n>a_{n+1}$. Furthermore, as $f\circ f$ is increasing, the subsequences of even terms and odd terms $(a_{2n})$ and $(a_{2n+1})$ are monotonic, and have different monotonicity.
So $(a_{2n})$ is increasing if and only if $a_0<a_2$, i.e. $$a_0<\frac{3a_0+4}{2a_0+3}\iff 2a_0^2< 4\iff 0<a_0<\sqrt 2.$$ One also checks that $a_0<a_1\iff a_o<\sqrt 2$, and $a_1>\sqrt 2$. As a result
So in any case, both sequences are adjacent sequences, which converge to the same limit. As a conclusion $(a_n)$ converges to $\sqrt 2$.
Note: not obvious, to get the matrix for a composition of Möbius transformations you multiply the matrices, in the same order.
The problem is largely about the Pell equations $ p^2 - 2 q^2 = \pm 1.$ You have an integer matrix $$ M = \left( \begin{array}{rr} 1 & 2 \\ 1 & 1 \end{array} \right) $$ so $$ M^2 = \left( \begin{array}{rr} 3 & 4 \\ 2 & 3 \end{array} \right) $$ and $$ M^3 = \left( \begin{array}{rr} 7 & 10 \\ 5 & 7 \end{array} \right) $$ and $$ M^4 = \left( \begin{array}{rr} 17 & 24 \\ 12 & 17 \end{array} \right) $$ Let us name $$ M^n = \left( \begin{array}{rr} p_n & 2 q_n \\ q_n & p_n \end{array} \right). $$ Various things happen, because we are dealing with the continued fraction for $\sqrt 2.$ For example, $$ p_n^2 - 2 q_n^2 = ( -1)^n. $$ Furthermore $$ p_{n+2} = 2 p_{n+1} + p_n, $$ $$ q_{n+2} = 2 q_{n+1} + q_n. $$ Here we have $$ p_1 = 1, p_2 = 3, p_3=7, p_4 = 17, $$ $$ q_1 = 1, q_2 = 2, q_3 = 5, q_4 = 12, $$ Note that both $p,q$ got to infinity, both can be written as a constant times $(1 + \sqrt 2)^n$ plus some other constant times $(1 - \sqrt 2)^n,$ the former larger than $1$ and the latter between $-1$ and $0.$
We do not need much specific other than $$ \frac{p^2}{q^2} - 2 = \frac{(-1)^n}{q^2} $$ so that the limit of $p/q$ really is $\sqrt 2.$ Here I begin to drop the subscript $n$'s on the $p,q$ numbers.
With initial value $a > 0,$ your fraction is the ratio of the entries in $$ \left( \begin{array}{rr} p_n & 2 q_n \\ q_n & p_n \end{array} \right) \left( \begin{array}{r} a \\ 1 \end{array} \right) \mapsto \frac{a p_n + 2 q_n}{a q_n + p_n}, $$ or $$ \frac{a p + 2 q}{a q + p} $$ as an abbreviation. Not as much trickery needed as I had first thought. From here we get $$ \frac{a p + 2 q}{a q + p} = \frac{a \frac{p}{q} + 2 }{a + \frac{p}{q}} = \frac{a \frac{p}{q} + \frac{p^2}{q^2} }{a + \frac{p}{q}} + \frac{2 - \frac{p^2}{q^2}}{a + \frac{p}{q}} = \frac{p}{q} + \frac{ \frac{(-1)^{n+1}}{q^2}}{a + \frac{p}{q}} $$ so that $$ \frac{a p + 2 q}{a q + p} = \frac{p}{q} \; + \; \frac{ (-1)^{n+1}}{a q^2 + pq} $$ That finishes it. Note that $$ \left| \frac{p}{q} - \sqrt 2 \right| < \frac{1}{q^2}. $$ With $$ q_n = \frac{1}{2 \sqrt 2} \left(1 + \sqrt 2 \right)^n - \frac{1}{2 \sqrt 2} \left(1 - \sqrt 2 \right)^n $$ we see that the distance from $\sqrt 2$ becomes quite small. Meanwhile $$ p_n = \frac{1}{2 } \left(1 + \sqrt 2 \right)^n + \frac{1}{2 } \left(1 - \sqrt 2 \right)^n $$
Hint
Show that $(a_{2n})$ and $(a_{2n+1})$ are monotonic and bounded (thus convergent). Compute there potential limit (you'll be surprised ;-)), and then conclude.
