Cauchy sequence of positive terms, find it's limit.

Question

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.

Answer 1

$|x_{n+2}-x_{n+1}| \leq 0.2|x_{n+1}-x_n| \Rightarrow |x_{n+k}-x_{n}|\leq c^k |x_2-x_1|=c^k$

$s=\frac{1}{|c|}-1 \Rightarrow |c|=\frac{1}{1+s}$ 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

$\lim_{n \to \infty}x_n=a \Rightarrow \lim_{n \to \infty}x_{n+1}=a$

$x_{n+1}=\frac{3(1+x_n)}{3+x_n} \Rightarrow \lim_{n \to \infty}x_{n+1}=\lim_{n \to \infty}\frac{3(1+x_n)}{3+x_n}=\frac{3a+3}{3+a}=a$

$\frac{3a+3}{3+a}=a \Rightarrow a= \sqrt3$

Answer 2

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 $a=f(a)$.

Answer 3

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 $$ \lim_{n\to\infty}x_n=\sqrt3. $$