analysis - 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?

Answer

We already proved (in Show limit for recursive sequence by induction) that $a_n\geq {1\over 3}$ so
$$a_{n+1}-a_n=a_n{1-3a_n\over 3a_n+3}\leq 0$$
so the sequence is decreasing and bounded so it is convergent. Say $a$ is it limit, then
\begin{eqnarray*}
a&=&\lim _{n\to \infty} a_{n+1} \\
&=& \lim _{n\to \infty} {4a_n\over 3a_n+3}\\
&=& {\lim _{n\to \infty}4a_n\over \lim _{n\to \infty}(3a_n+3)}\\
&=& {4a\over 3a+3}

\end{eqnarray*}
So we have to solve the equation $$a={4a\over 3a+3}\Longrightarrow 3a^2-a=0 \Longrightarrow a=0\;\; {\rm or}\;\; a=1/3$$
Since all members of sequence are $\geq {1/3}$ we have $$a ={1\over 3}$$