Convergence of a sequence $a_{n+1}=\frac{2(2a_n+1)}{a_n+3},n=1,2,...,a_1=1$ I have simply checked first five terms from where it is obvious that its limit is $L=2$, thus it is convergent sequence.
I am interested in how to prove by induction that sequence is bounded and is it necessary?
Thanks for replies. 
 A: Clearly the sequence is positive for all $\;n\in\Bbb N\;$ . First:
$$a_1=1\le 2\implies a_{n+1}=\frac{2(2a_n+1)}{a_n+3}\le 2\iff 4a_n+2\le 2a_n+6\iff$$
$$\iff 2a_n\le4\iff a_n\le 2\;,\;\;\text{and this proves by induction that}\;\;a_n\le 2\;\;\forall\,n\in\Bbb N$$
Also:
$$a_2=\frac{2(2\cdot1+1)}{1+3}=\frac32\ge1=a_1$$
Induction:
$$a_{n+1}=\frac{2(2a_n+1)}{a_n+3}\ge a_n\iff 4a_n+2\ge a_n^2+3a_n\iff a_n^2-a_n-2\le0\iff$$
$$\iff (a_n-2)(a_n+1)\le 0\iff -1\le a_n\le 2$$
and the last double inequality is true by the first part. Thus, our sequence is bounded and monotonically ascending and thus it has a limit, call it $\;\alpha\;$ . Now, using arithmetic of limits:
$$\alpha=\lim_{n\to\infty}a_{n+1}=\frac{2(2\alpha+1)}{\alpha+3}\implies \alpha=2$$
A: Here is another way to approach this problem: The map
$$T(x):={2(2x+1)\over x+3}$$
is a Moebius transformation whose fixed points compute to $x_1=2$ and $x_2=-1$. We introduce a new coordinate variable $u$ such that these two points correspond to $u_1=0$ and $u_2=\infty$. This means that we define $u$ by
$$u:={x-2\over x+1}=:S(x)\ .$$
It follows that $S^{-1}$ is given by
$$x=S^{-1}(u)={2+u\over1-u}\ .$$
The given $T$ then appears in terms of $u$ as
$$\hat T(u)=S\circ T\circ S^{-1}(u)={2\over 5} u\ ,$$
and one has $u_1=S(a_1)=-{1\over2}$. It is now obvious that $$\lim_{n\to\infty}\hat T^n(u_1)=0\ ,$$
and this implies
$$\lim_{n\to\infty}T^n(a_1)=S^{-1}(0)=2\ .$$
A: If $0<a_n<2$ then  $a_n+3>0$ and
$$2a_n<4$$
$$4a_n+2<2a_2+6$$
$$2(2a_n+1)<2(a_n+3)$$
$$\frac{2(2a_n+1)}{(a_n+3)}<2$$
$$0<a_{n+1}<2$$
Since $a_1=1<2$ all the $a_n$ are bounded above by $2$.
