Show convergence of recursive sequence and find limit value Let $(a_n)_{n \in \mathbb N}$ be a recursive sequence. It is defined as $a_1=1, \quad a_{n + 1} = \frac{4a_n}{3a_n+3}$. 
I have to show that the sequences converges and find a limit value. 
To show convergence I was about to use the Cauchy criterum. Unfortunately I am quite confused here because of the recursive definition. 
Question: How can I show show that the sequences converges and how can I find a limit value?

If we write $a_n$ as $p_n/q_n$, we get:
$$a_{n + 1} = \frac{4(p_n/q_n)}{3(p_n/q_n)+3} = \frac{4p_n}{3p_n+3q_n} = \frac{p_{n+1}}{q_{n+1}}$$
In other words:
$$p_{n+1} = 4p_n \longrightarrow p_n = 4^{n-1}$$
$$q_{n+1} = 3p_n + 3q_n = 3\cdot 4^{n-1} +3 q_n$$
We can find $q_n$ with generating function $Q(x) = \sum_{n=1}^\infty q_nx^n$:
$$Q(x) = x + x\sum_{n=1}^\infty q_{n+1}x^n$$
$$Q(x) = x + 3x\sum_{n=1}^\infty (q_n + 4^{n-1})x^n$$
$$Q(x) = x + 3xQ(x) + 3x\sum_{n=1}^\infty 4^{n-1}x^n$$
$$(1 - 3x)Q(x) = x + 3x\frac{x}{1-4x}$$
$$Q(x) = \frac{x}{1-3x} + \frac{3x^2}{(1-3x)(1-4x)} = \frac{1}{4} +\frac{x}{1-3x} - \frac{1}{1-3x} + \frac{3}{4(1-4x)}$$
$$Q(x) = \frac{1}{4} + \sum_{n=1}^\infty3^{n-1}x^n - \sum_{n=1}^\infty3^{n}x^n + \frac{3}{4}\sum_{n=1}^\infty 4^nx^n$$
$$Q(x) = \frac{1}{4} + \sum_{n=1}^\infty(3^{n-1}  - 3^n + \frac{3}{4}4^n)x^n = \sum_{n=1}^\infty q_nx^n$$
$$q_n = 3^{n-1}  - 3^n + \frac{3}{4}4^n = \frac{1}{12}(9\cdot4^{n} - 8\cdot 3^{n})$$
Thus $a_n = \dfrac{3\cdot4^{n}}{9\cdot4^{n} - 8\cdot 3^{n}}$ giving limit $\dfrac{1}{3}$.
