Cauchy sequence of positive terms, find it's limit. Consider the sequence of positive terms, where $x_n$ it's difined by
\begin{equation}
x_1=3\ and\ x_{n+1}=\frac{3(1+x_n)}{3+x_n}.
\end{equation}
Find   $a=\underset{n\rightarrow \infty}\lim x_n$.
All that I have found is that exists $c=0.2$ such that
\begin{equation}
|x_{n+2}-x_{n+1}| \leq 0.2|x_{n+1}-x_n|.
\end{equation}
I also tried to prove that
\begin{equation}
x_{k+1} \leq x_k, 
\end{equation}
but I failed.
 A: Note that $x_{n+1}=f(x_n)$ where
$$ f(x)=\frac{3(1+x)}{3+x}=3-\frac6{x+3}.$$
We have $f(x)>3-\frac63>0$ for all $x\ne -3$.
Now if $x>y>0$ then $x+3>y+3$, then $\frac1{x+3}<\frac1{y+3}$, then $-\frac1{x+3}>-\frac1{y+3}$, and ultimately $f(x)>f(y)>0$. Therefore, $x_1=3>2=x_2$ implies by induction that $x_n>x_{n+1}>0$ for all $n$, i..e, the sequence is strictly decreasing and bounded from below, hence convergent to some $a$. For this $a$, we must have $a=f(a)$.
A: Clearly $x_n\ge0$ for all $n\in N$. Let $y_n=x_n-\sqrt3$ and then $y_n+\sqrt3\ge0$. Note
$$ |y_{n+1}|=\bigg|\frac{(3-\sqrt3)y_n}{3+\sqrt3+y_n}\bigg|\le\frac{3-\sqrt3}{3}|y_n| $$
and hence
$$ |y_n|\le\bigg(\frac{3-\sqrt3}{3}\bigg)^{n-1}|y_1|. $$
So
$$ \lim_{n\to\infty}y_n=0 $$
or
$$ \lim_{n\to\infty}x_n=\sqrt3. $$
A: $|x_{n+2}-x_{n+1}| \leq 0.2|x_{n+1}-x_n| \Rightarrow |x_{n+k}-x_{n}|\leq c^k |x_2-x_1|=c^k$
$s=\frac{1}{|c|}-1 \Rightarrow |c|=\frac{1}{1+s}$ and $Ns \geq \epsilon$
$|c|=\frac{1}{1+s} \Rightarrow |c|^k=(\frac{1}{1+s})^k=\frac{1}{(1+s)^k} \leq $
$\frac{1}{1+sk} < \frac{1}{sk} < \epsilon $
which means that $\lim_{n \to \infty}x_n=a$ exists
$\lim_{n \to \infty}x_n=a \Rightarrow \lim_{n \to \infty}x_{n+1}=a$
$x_{n+1}=\frac{3(1+x_n)}{3+x_n} \Rightarrow \lim_{n \to \infty}x_{n+1}=\lim_{n \to \infty}\frac{3(1+x_n)}{3+x_n}=\frac{3a+3}{3+a}=a$
$\frac{3a+3}{3+a}=a \Rightarrow a= \sqrt3$
