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'.

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 $\frac{1}{3}$.
