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 $n$.
 A: 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 $a_1=\frac 13$.
A: Define the function
$$
f(a)=\frac1{4-3a}\tag1
$$
Note that
$$
\begin{align}
f(a)-a
&=\frac{(3a-1)(a-1)}{4-3a}\tag{2a}\\
&\left\{\begin{array}{}
\lt0&\text{if }a\in\left(\frac13,1\right)\cup\left(\frac43,\infty\right)\\
\gt0&\text{if }a\in\left(-\infty,\frac13\right)\cup\left(1,\frac43\right)
\end{array}\tag{2b}
\right.
\end{align}
$$
Consider the two sequences for $n\in\mathbb{Z}$,
$$
\begin{align}
p_n
&=\frac{3^{n-1}+1}{3^n+1}\tag{3a}\\
&=\frac13\left(1+\frac2{3^n+1}\right)\tag{3b}
\end{align}
$$
and
$$
\begin{align}
q_n
&=\frac{3^{n-1}-1}{3^n-1}\tag{4a}\\
&=\frac13\left(1-\frac2{3^n-1}\right)\tag{4b}
\end{align}
$$
where $q_0=\pm\infty$.
Note that
$$
\begin{align}
f(p_n)&=p_{n+1}\tag{5a}\\
f(q_n)&=q_{n+1}\tag{5b}
\end{align}
$$
where, in the case of $q_0$,
$$
\begin{align}
f(q_{-1})&=f\!\left(\tfrac43\right)=\infty=q_0\tag{6a}\\
f(q_0)&=f(\infty)=0=q_1\tag{6b}
\end{align}
$$
Define the intervals
$$
\begin{align}
P_n&=(p_{n+1},p_n)\tag{7a}\\
Q_n&=(q_n,q_{n+1})\tag{7b}
\end{align}
$$
where $Q_{-1}=\left(\frac43,\infty\right)$ and $Q_0=\left(-\infty,0\right)$:

In the animation above, the solid red and green lines are the $P_n$ and $Q_n$. The arrows point to the dotted intervals $P_{n+1}$ and $Q_{n+1}$. The intervals are red if $f(a)\lt a$ on that interval and green if $f(a)\gt a$; these intervals are described in $(2)$.
Since $f'(a)\gt0$ except at $q_{-1}=\frac43$ (which is between $Q_{-2  }$ and $Q_{-1}$), we have the bijections
$$
\begin{align}
f&:P_n\to P_{n+1}\tag{8a}\\
f&:Q_n\to Q_{n+1}\tag{8b}
\end{align}
$$
Since
$$
\bigcup_{n\in\mathbb{Z}}P_n\cup\bigcup_{n\in\mathbb{Z}}Q_n\cup\left\{p_n:n\in\mathbb{Z}\right\}\cup\left\{q_n:n\in\mathbb{Z}^{\ne0}\right\}=\mathbb{R}\tag9
$$
$(5)$ and $(8)$ show that for all points except $\left\{q_n:n\le0\right\}\cup\{1\}$, iterating $f$ will produce a seqence converging to $\frac13$ (one might even say that $q_{-\infty}=1$).
A: Hint: If $a_1<1$, it is easy to see $a_n<1$ and then let $b_n=a_n-\frac13$.
If $a_1\in(1,\frac43)$, it is easy to see $a_n\in(1,\frac43)$ and then let $b_n=a_n-1$. You can do the rest.
A: Without induction.
If you follow the steps described here which I used answering this question making the story short
$$a_{n+1}=\frac{1}{4-3 a_{n}} \qquad \text{with} \qquad a_1=c$$
$$a_n=\frac 13\frac{c \left(3^n-9\right)-(3^n-3) } {c(3^n-3)-(3^n-1) }$$
Now, you need to consider the various cases to arrive to the nice results from  @Brian M. Scott's nice analysis.
