Show that the sequence $a_1=3$ and $a_{n+1}=\frac{3(1+a_n)}{3+a_n}$ converges to $\sqrt3$ First of all let´s see how it goes:  $3,\frac{3(1+3)}{3+3}=\frac{12}{6}=2,\frac{3(1+2)}{3+2}=\frac{9}{5}=1.8,...$ We see that it's decreasing.
What i need to show first is that the sequence decreases. To do that, i´d say:
$$a_n-a_{n+1}>0\Rightarrow a_n-{\frac{3(1+a_n)}{3+a_n}}>0\Rightarrow a_n^2>3\Rightarrow a_n>\sqrt{3}$$ But i think this is like say that $a_n$ converges to $\sqrt3$, which is asked to be proved. So, How can i show this sequence decreases?.  And to show it's bounded?

We have $a_1>0$ and from the definition of $a_{n+1}$ we have $a_n>0\implies a_{n+1}>0 $ for any $n$. So by induction we have $a_n>0$ for all $n\in \Bbb N.$
We have $a_1>\sqrt 3$. If $a_n>\sqrt 3$ then, since $a_{n+1}>0,$ we have  $$a_{n+1}>\sqrt 3\iff a_{n+1}^2>3\iff$$ $$\iff \frac {3^2(1+a_n)^2}{(3+a_n)^2}>3\iff$$ $$\iff 3^2(1+a_n)^2 -3(3+a_n)^2>0\iff$$ $$\iff 6(a_n^2-3)>0.$$ So by induction we have $a_n>\sqrt 3$ for all $n\in \Bbb N.$
Since $3+a_n>a_n>0$ we have $$a_n>a_{n+1} \iff (3+a_n)\left(a_n-\frac {3(1+a_n)}{3+a_n}\right)>0 \iff$$ $$\iff 3a_n+a_n^2> 3+3a_n\iff a_n^2>3,$$ and we  have already shown that $a_n^2>3.$ So $a_n>a_{n+1}$ for all $n.$
Since $(a_n)_n$ is a decreasing sequence of positive terms  it has a limit $L\geq 0$ . And we have $$L=\lim_{n\to \infty}a_{n+1}=\lim_{n\to \infty}\frac {3(1+a_n)}{3+a_n}= \frac {3(1+L)}{3+L}.$$ So $L=\frac {3(1+L)}{3+L}, $ 
 which implies $L^2=3.$ Now $(L\geq 0\land L^2=3)\implies L=\sqrt 3.$
