Application of Induction in the analyze of the convergence a sequence defined recursive. 
Let $\left\{a_{n}\right\}$ be defined recursively by
$$
a_{n+1}=\frac{1}{4-3 a_{n}}, \quad  n \geq 1
$$
Determine for which $a_{1}$ the sequence converges and in case of convergence find its limit.

My approach: Note that  $$a_{n +1}=\frac{1}{4-3a_{n}}, \quad n\geq 1$$
so, firstly  I would like to find $a_{n}$. Now, I was trying to find a  pattern but I can't find this
\begin{eqnarray*}
n=1 &\implies & a_{2}=\frac{1}{4-3a_{1}}=\frac{(3^{2-1}-1)-(3^{2-1}-3)a_{1}}{(2^{2})-(3^{2}-6)a_{1}}\\
n=2 & \implies & a_{3}=\frac{1}{4-3a_{2}}=\frac{1}{4-3 \left( \frac{1}{4-3a_{1}}\right)}=\frac{4-3a_{1}}{4(4-3a_{1})-3}\\
\vdots &\implies & \vdots \\
\end{eqnarray*}
If I know $a_{n}=a_{n}(a_{1})$, so I can analyze the denominator for the conclude when $a_{n}$ is not defined.
How can find $a_{n}$?
Also I know this problem was answered here. But I think, we can find an elementary solution using induction on $n$.

Update: Thanks Brian M. Scott for your insight.
I'll add the case where some $a_k=\frac 43$. Per Brian, we need to solve for the sequence $b_k$ such that $b_1=\frac 43$, $b_{k+1}=\frac{4b_k-1}{3b_k}$. This can be solved in a similar fashion, but easier because $b_1$ is given.
Note that
$$
b_{k+1} - 1 = \frac{b_k-1}{3b_k}$$$$
b_{k+1} - \frac 13 = \frac{b_k-\frac{1}{3}}{b_k}\tag 1
$$
From $(1)$ we conclude $b_k>\frac 13, \forall k$ via induction.
Then $\frac{b_{k+1}-1}{b_{k+1}-\frac 13} = \frac{1}{3} \frac{b_k-1}{b_k-\frac 13} \implies \frac{b_k-1}{b_k-\frac 13} = \frac{1}{3^{k-1}} \left( \frac{b_1 - 1}{b_1 - \frac 13}\right) = \frac{1}{3^k}$
Therefore $b_k = \frac{1 - \frac{1}{3^{k+1}}}{1-\frac{1}{3^l}} = \frac{3^{k+1} -1}{3^{k+1}-3}$ which is the same as Brian's results.

Original answer:
Since $1$ and $\frac 13$ are roots of the characteristic equation $x=\frac{1}{4-3x}$, we have
$$a_{n+1}-1 = \frac{3(a_n-1)}{4-3a_n}$$
$$a_{n+1}-\frac 13 = \frac{a_n-\frac 13}{4-3a_n}$$
So if no $a_n = \frac 13$ you have
$$\frac{a_{n+1}-1}{a_{n+1}-\frac 13} = 3 \frac{a_n-1}{a_n-\frac 13} = 3^n \frac{a_1-1}{a_1-\frac 13}$$
Of course you need to take care of the case where $a_1=\frac 13$.
