Show that $\{a_n\}$ defined by $a_{n+1}=\frac{a_n+2}{a_n+1}$ converges 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.
 A: 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 $0$.
A: 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


*

*if $a_0<\sqrt 2$, one has
$$a_0<a_2<\dots<a_{2n}<\dots<\sqrt 2<\dots<a_{2n+1}<\dots<a_1. $$
This proves $(a_{2n}$ is bounded from above and $(a_{2n+1})$ is bounded from below. By the monotonic convergence theorem, both converge. Further, as they're defined by the same recurrence $u_{n+2}=\dfrac{3u_n+4}{2u_n+3}$, they converge to a positive fixed point of the function $\dfrac{3x+4}{2x+3}$. The only such point is $\sqrt2$.

*if $a_0>\sqrt 2$, one has
$$a_1<\dots<a_{2n+1}<\dots<\sqrt 2<\dots<a_{2n}<\dots<a_0.$$
The rest of the computations is the same.


So in any case, both sequences are adjacent sequences, which converge to the same limit. As a conclusion $(a_n)$ converges to $\sqrt 2$.
A: 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
$$ p_n = \frac{1}{2 } \left(1 + \sqrt 2 \right)^n + \frac{1}{2 } \left(1 - \sqrt 2 \right)^n    $$
A: 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.
