Find $a_1$ so that $ a_{n+1}=\frac{1}{4-3a_n}\ ,n\ge1 $ is convergent Let $  \left\{ a_n \right\}  $ be a recursive sequence such that $$a_{n+1}=\frac{1}{4-3a_n}\quad,n\ge1  $$
Determine for which $a_1$ the sequence converges and in case of convergence find its limit. 
The problem is from the book 'Problems in Mathematical Analysis I by W.J.Kaczor'.
 A: The Moebius transformation
$$T:\quad\bar{\mathbb C}\to\bar{\mathbb C},\qquad x\mapsto T(x):={1\over 4-3x}$$
has the two fixed points $1$ and ${1\over3}$. We therefore introduce a new complex projective coordinate $z$ via
$$z:={x-{1\over3}\over x-1},\qquad{\rm resp.},\qquad x={z-{1\over3}\over z-1}\ .$$
In terms of this coordinate $T$ appears as ${\displaystyle \hat T(z)={z\over3}}$ (with fixed points $0$ and $\infty$), so that
$$\bigl(\hat T\bigr)^{\circ n}(z)={z\over 3^n}\ .$$
It follows that for all initial points $z\ne\infty$ we have $$\lim_{n\to\infty}\bigl(\hat T\bigr)^{\circ n}(z)=0\ .$$
In terms of the original variable $x$ this means that for all initial points $x\ne1$ we have
$$\lim_{n\to\infty}T^{\circ n}(x)={1\over3}\ .$$
There is, however, the following caveat: The above argument refers to the domain $\bar{\mathbb C}$; but maybe you want to exclude $x=\infty$ as a generic point. In terms of the coordinate $z$ this is the point $z_*=1$. For all initial values $z_k=3^k$ $(k\geq1)$ we have $\bigl(\hat T\bigr)^{\circ k}z_k=z_*$. This implies that in the original formulation of the problem you have $T^{\circ k}(x_k)=\infty$ (i.e., you "accidentally" hit $\infty$ after finitely many steps) for all initial points  $x_k=\bigl(3^k-{1\over3}\bigr)/(3^k-1)$ $(k\geq1)$.
A: Suppose that this sequence does converge to A.  Then we must have $A= \frac{1}{4- 3A}$.  Then $A(4- 3A)= 4A- 3A^2= 1$.  $3A^2- 4A+ 1= (3A- 1)(A- 1)= 0$. A is either 1 or 1/3.  
If $a_1> 1$ the sequence clearly converges to 1.  If $a_1\le 1/3$ if clearly converges to $\frac{1}{3}$.  It's a little harder to show, but still true, that if $1/3< a_1< 1$ then the sequence converges to 1/3: if $\frac{1}{3}< a< 1$ then $1< 3a< 3$ so that 0< 3a-1< 2.  But for $\frac{1}{3}< a< 1$, $a- 1< 0$.  That is, 3a-1 is positive while a- 1 is negative so that $(3a- 1)(a- 1)= 3a^2- 4a+ 1< 0$.  Then $3a^2- 4a= a(3a- 4)> 1$ and $a> \frac{1}{3a- 4}$.  That is, for $\frac{1}{3}< a_1< 1$ the sequence is decreasing to $\frac{1}{3}$.
