Limit of sequence $a_{n+1}=3/(2+a_n)$, $a_1=0$ The task is to find the limit of sequence, given by following recursion:
$$ a_1=0, a_{n+1}=\frac{3}{2+a_n} $$
so at first I tried to find some first parts of the sequence
$$a_1=0, a_2=\frac{3}{2}, a_3=\frac{6}{7}, a_4=\frac{21}{20}, a_5=\frac{60}{61}, a_6=\frac{183}{182}$$
Maybe I done some kind of mistake in these calculations, but nevertheles It seems like this sequence has a limit equal to 1, and approches it from left for odd indexes $$a_1, a_3, a_5...$$ 
and from right for even indexes
$$a_2, a_4, a_6...$$ 
Now I had idea to split this sequence to two subsequences, one for odd, and one for even indexes, but one is still connected to another and I got stuck in nowhere. Any clue, how to deal with this task, whitout genereting functions? 
Any kind of help would be appriciated.
(PS. I am first year student of maths)
Update:
I found that even indexed ones can be described by:
$$a_{2n+2}=\frac{6+3a_{2n}}{7+2a_{2n}} $$
And similary odd one's
$$a_{2n+1}=\frac{6+3a_{2n-1}}{7+2a_{2n-1}} $$
And from that I got Did's hint...(checked).
 A: Following @Crostul's hint, consider the function $f(x) = \frac 3{x+2}$ for $x\geqslant 0$. Then for $x,y\geqslant 0$ we have
$$|f(x)-f(y)| = 3\left|\frac{x-y}{(x+2)(y+2)}\right| \leqslant \frac34|x-y|, $$
so $f$ is a contraction mapping. It follows from the Banach fixed-point theorem that $f$ has a unique fixed point. By inspection, we see that
$$ f(1) = \frac3{1+2}=1. $$
Now, as clearly each $a_n\geqslant 0$, we have $a_{n+1}=f(a_n)$. From the Lipschitz condition we see that $$|f(a_n)-f(1)|=|a_{n+1}-1|\leqslant \frac34|a_n-1|, $$
and hence $$\left|\frac{a_{n+1}-1}{a_n-1} \right|\leqslant \frac34.$$
By induction we may show that
$$\left|\frac{a_{n+m}-1}{a_n-1}\right|\leqslant\left(\frac34\right)^m, $$
and hence
$$ |a_{n+m}-1|\leqslant\left(\frac34\right)^m|a_n-1|.$$ 
It follows that $$\lim_{m\to\infty}|a_{n+m}-1|=0.$$
As $n$ is arbitrary, we have
$$\lim_{n,m\to\infty}|a_{n+m}-1|=0, $$
from which we conclude that $$\lim_{n\to\infty}a_n=1.$$
A: We know that if a sequence $(a_n)$ has a limit $l$, then the sequence $(a_{n+1})$ also has the limit $l$. (You can verify this for your own convenience for any sequence).
Therefore $$\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}a_{n}=l$$
So we have $$a_{n+1}=\frac{3}{2+a_n}$$
$$\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\frac{3}{2+a_n}$$
$$\lim_{n\to\infty}a_{n+1}=\frac{3}{2+\lim_{n\to\infty}a_n}$$
$$l=\frac{3}{2+l}$$
$$l^2+2l-3=0$$
$$(l+3)(l-1)=0$$
$$l=1 \,\ \text{or} \,\ -3$$
But $l$ cannot be negative since $a_1=0$ and $(a_n)$ is a sequence of positive terms oscillating about $1$.
So $$\lim_{n\to\infty}a_{n}=l=1$$
EDIT:
Take the positive terms $1,1$ and $a_n$. So by A.M-G.M. inequality, we have 
$$\frac{1+1+a_n}{3}\ge (a_n)^{\frac{1}{3}}$$
$$\frac{3}{2+a_n}\le \frac{1}{(a_n)^{\frac{1}{3}}}$$
$$a_{n+1}\le (\frac{1}{a_n})^{\frac{1}{3}}$$
$$a_{n}\le (\frac{1}{a_{n-1}})^{\frac{1}{3}}$$
Now you can check that $min(a_n)=\frac{6}{7}$ omitting $a_1=0$.
Hence maximum value of $\frac{1}{a_n}=\frac{7}{6}$
So $$a_{n+1}\le (\frac{7}{6})^{\frac{1}{3}}$$
This shows the sequence is bounded.
I don't know but see if you can use this to prove Cauchy.
A: This is a pure algebraic approach which map the non-linear recurrence relation
to a matrix valued linear recurrence relation.
Writing $a_n$ as $\displaystyle\;\frac{p_n}{q_n}$, we have:
$$a_{n+1} = \frac{3}{2+a_n}\iff 
\begin{bmatrix}p\\ q\end{bmatrix}_{n+1} \propto A \begin{bmatrix}p\\ q\end{bmatrix}_n
\quad\text{ where }\quad A = \begin{bmatrix}0 & 3\\ 1 & 2\end{bmatrix} $$
Together with $\displaystyle\;a_1 = 0 \implies \begin{bmatrix}p \\ q\end{bmatrix}_1 \propto \begin{bmatrix}0 \\ 1\end{bmatrix}$, this implies
$$\begin{bmatrix}p\\ q\end{bmatrix}_n \propto A^{n-1} \begin{bmatrix}0\\ 1\end{bmatrix}$$
Notice the characteristic polynomial for $A$ is
$$\det\left(\lambda I_2 - A\right) = \lambda (\lambda - 2) - 3 = (\lambda+1)(\lambda-3)$$
By Cayley-Hamilton theorem, we have
$$(A + I_2)(A - 3I_2) = 0\;\implies\;
\begin{cases}
A^k (A+I_2 ) &= 3^k (A+I_2)\\
A^k (A- 3I_2)&= (-1)^k(A-3I_2)
\end{cases}\quad\text{ for all } k \ge 0$$
Using the partition of identity $$I_2 = \frac14\left((A+I_2) - (A -3I_2)\right),$$
we find
$$A^k = \frac14 A^k\left((A+I_2) - (A -3I_2)\right) = 
\frac14\left(3^k (A+I_2) - (-1)^k (A-3I_2)\right)$$
and hence
$$\begin{bmatrix}p\\ q\end{bmatrix}_n \propto \left( 3^{n-1}(A+I_2) - (-1)^{n-1} (A-3I_2)\right)
\begin{bmatrix}0\\ 1\end{bmatrix}
= 3^n \begin{bmatrix}1\\ 1\end{bmatrix} - (-1)^{n-1}\begin{bmatrix}3\\ -1\end{bmatrix}$$
This implies
$$a_n = \frac{p_n}{q_n} = \frac{3^n + 3(-1)^n}{3^n - (-1)^n}
= \frac{1 + 3(-\frac13)^n}{1 - (-\frac13)^n}
$$
Since both the numerator and denominator converges to $1$ as $n \to \infty$. $a_n$ also converges to $1$.
A: Using the advanced hint from Did, 
$$\begin{align}
\frac{a_n-1}{a_n+3}
&=\left(-\frac 13\right)\frac {a_{n-1}-1}{a_{n-1}+3}\\
&=\left(-\frac 13\right)^2\frac {a_{n-2}-1}{a_{n-2}+3}\\
&\vdots\\
&=\left(-\frac 13\right)^{n-1}\frac {a_0-1}{a_0+3}\qquad\qquad\qquad(a_0=0)\\
&=\left(-\frac 13\right)^{n}\underset{\small{\text{as  } n\to \infty}}{\longrightarrow} 0\\
a_n-1&\underset{\small{\text{as  } n\to \infty}}{\longrightarrow} 0\\
a_n&\underset{\small{\text{as  } n\to \infty}}{\longrightarrow} 1\quad\blacksquare
\end{align}$$

NB: The advanced hint from Did may be derived by writing the original recurrence relation in the form
$$\frac {a_{n+1}}1=\frac 3{2+a_n}=\frac TB$$
and by componendo and dividendo,
$$\begin{align}\frac {T-B}{T+3B}:\qquad \qquad \frac {a_{n+1}-1}{a_{n+1}+3}&=\frac {3-(2+a_n)}{3+3(2+a_n)}\\
&=-\frac13\cdot \frac {a_n-1}{a_n+3}\qquad\qquad\qquad\qquad \end{align}$$
