Prove $a_1 = 1, a_{n+1} = \frac{1+a_n}{2+a_n}$ converges I need to prove that the sequence defined by
$$a_1 = 1, a_{n+1} = \frac{1+a_n}{2+a_n}$$
converges.
I tried to prove that it's bounded and monotonically decreasing, but I couldn't prove it's monotonically decreasing.
I also managed to find the limit assuming it converges.
 A: Note that 
$$a_{n+1} = \frac{1+a_n}{2+a_n} = 1 - \frac{1}{2 + a_{n}}.$$
So, given that both $a_n$ and $a_{n-1}$ are positive, we have
$$
a_{n} < a_{n-1} \implies \frac {1}{2 + a_n} > \frac 1{2 + a_{n-1}}
\implies 1 - \frac {1}{2 + a_n} < 1 - \frac 1{2 + a_{n-1}} \implies
a_{n+1} < a_n.
$$ 
So, we can indeed conclude that the sequence is monotonic.
A: Rewrite the recurrence relation as  $$a_{n+1} = \frac{1+a_n}{2+a_n} = 1-\frac1{2+a_n}$$
We prove by induction that $(a_n)_n$ is monotonically decreasing. Clearly $$a_1 = 1 > \frac23 = a_2.$$
Assume that $a_n \ge a_{n-1}$. Then also 
$$a_{n+1} = 1-\frac1{2+a_n} \ge 1-\frac1{2+a_{n-1}} = a_n$$
A: To prove it converges, you can prove by induction that $a_{n+1} \le a_{n} \; \forall n \in \Bbb N^* $ and that it is lower bounded by a fixed point of $g(x):={1+x\over 2+x}$.
Therefore it's decreasing and converges to $l \in \Bbb R$, so that $l={1+l\over 2+l}$.
A: Let $f(x) = {1+x \over 2+x}$. Note that $f([0,1]) \subset [0,1]$. Also note that
$f'(x) = {1 \over (2+x)^2} < {1 \over 4}$, so $f$ is a contraction.
Hence $f$ has a unique fixed point given by the limit of the sequence $a_{n+1} = f(a_n)$.
From continuity we see that $a = f(a) $ and so it solves $a^2+a-1=0$ and the only solution that lies in $[0,1]$ is $a={1 \over 2}(\sqrt{5}-1)$.
