Convergence of sequence $a(n+1) = \frac{3(1+a(n))}{3+a(n)}$ The sequence is defined by: 
$a(1)=1, a(n+1) = \frac{3(1+a(n))}{3+a(n)}$
How do I show that the sequence is monotonic and bounded from above?
My approach to show its monotone was:
$a(n+1) - a(n) \geq 0$ 
$ \frac{3(1+a(n))}{3+a(n)} - a(n) = \frac{3+3a(n)}{3+a(n)} - \frac{3a(n)+ a(n)^2}{3+a(n)} = \frac{3(1+a(n)) - a(n)(3+a(n))}{3+a(n)} = \frac{3-a(n)^2}{3+a(n)} \geq  ? \geq 0$
I know the sequence is bounded from above by $\sqrt{3}$ but I'm totally lost at how to show this.

Notice that $$\dfrac{a_{n+1}-\sqrt{3}}{a_{n+1}+\sqrt{3}}=\dfrac{a_{n}-\sqrt{3}}{a_{n}+\sqrt{3}}\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$$. We can calculate the accurate value of $a_n$. 
