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?
Best 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 $$a ={1\over 3}$$