Prove that $a_{n+1}=3\frac{a_{n}+1}{a_{n}+3} , a_{1} =a> 0$ is convergent The sequence $a_{n}$ is given in this recursive form:
$$a_{n+1}=3\frac{a_{n}+1}{a_{n}+3}  , a_{1} =a> 0$$  
How does one show that it is convergent?
I tried to prove that it is monotone increasing and bounded by $\sqrt3$ but than I thought that it may be better to seperate into cases which $a>\sqrt3$ and $a<\sqrt3$,
Either way I did'nt get to a solution and wasn't able to write the proper Induction.
EDIT: please help me with only sequences theorems, without Lhopital and functions.
Thank you
 A: You are on he right track. Let $f(x)=3(x+1)/(x+3)$. It is a continuous, increasing function. It cuts the diagonal of the first quadrant (the line $y=x$) at only one point: $x=\sqrt3$, $y=\sqrt3$. If $0\le x<\sqrt3$, then $x<f(x)<\sqrt3)$; this is all you need to prove (by induction) that in this case $\{a_n\}$ is a bounded increasing sequence. On the other hand, if $x>\sqrt3$, then $\sqrt3<f(x)<x$; this implies that $\{a_n\}$ is decreasing and bounded below by $\sqrt3$. Once you know the sequence is convergent, taking limits on both sides of the recurrence relation shows that the limit must be a solution of the equation $f(x)=x$. Since the only positive solution is $x=\sqrt3$, this is the limit.
A: You are correct to divide the cases (there is also $a=\sqrt{3}$)  


*

*For $a_n =\sqrt{3}$, you should be
able to show $a_{n+1} =\sqrt{3}$  

*For $0 < a_n < \sqrt{3}$, you should
be able to show $a_n < a_{n+1} < \sqrt{3}$ 

*For $a_n > \sqrt{3}$, you
should be able to show $a_n > a_{n+1}  > \sqrt{3}$


So you either have a constant, or an increasing sequence bounded above or a decreasing sequence bounded below.  In any case you have a converging sequence.
If it helps, you have   $a_{n+1}=3 - \dfrac{6}{a_{n}+3} $ 
A: I will sketch half of a solution of the problem.  You can do the second half.
If $a$ is positive, all terms are obviously positive.  Suppose that $a<\sqrt{3}$. We show that the sequence is bounded and increasing.
First show that the sequence is bounded.  We show that if $a_n^2<3$, then $a_{n+1}^2 <3$.
Calculate:
$$3-a_{n+1}^2=3-\frac{9(a_n+1)^2}{(a_n+3)^2}=\frac{6(3-a_n^2)}{(a_n+3)^2}$$
so if $a_n^2<3$ then $a_{n+1}^2<3$.
Now show that the sequence is increasing. Note that
$$a_{n+1}-a_n=\frac{3(a_n+1)}{a_n+3}-a_n=\frac{3-a_n^2}{a_n+3}$$
By what was shown eaarlier, the right-hand side of the above equation is positive.
Now it's your turn.  Suppose that $a>\sqrt{3}$.  Continue.  
A: If the sequence has a limit $\alpha$ then $\alpha$ has to satisfy the equation $\alpha=3{\alpha+1\over\alpha+3}$, so $\alpha$ would have to be one of $\pm\sqrt3$, and as all $a_n$ are $> 0$, we are left with $\alpha=\sqrt3$. This suggests putting $a_n=\sqrt3+b_n$ $\ (n\geq 1)$ in the hope that we can prove $\lim_{n\to\infty} b_n=0$. From
$$\sqrt3 + b_{n+1}=3{\sqrt3 + b_n+1 \over \sqrt3 + b_n+3}$$
we get the following recursion formula for the $b_n$:
$$b_{n+1}=b_n \ {3-\sqrt3 \over \sqrt3 + b_n+3}= b_n\ {3-\sqrt3 \over a_n+3}\ .$$
Here the last fraction is positive and $< (3-\sqrt3)/3<{1\over2}$ from which it follows that the $b_n$ indeed converge to $0$.
