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 by induction, that it is limited for $a_n \ge \frac{1}{3}$.
So as an induction step I have $n \to n + 1$. And the idea is to get $a_{n+2}$ and show somehow $a_{n+2} \ge \frac{1}{3}$
$$a_{n+1} = \frac{4a_n}{3a_n + 3} $$
$$a_{n+2} = \frac{4 \frac{4a_n}{3a_n + 3}}{3 \frac{4a_n}{3a_n + 3} + 3} = \frac{ \frac{16a_n}{3a_n+3} }{ \frac{12 a_n}{3a_n + 3} + 3}$$
Question: How should I go on or am I on the wrong path?
Edit: You can find other answers for the exact same question here.
Best Answer
You wonder if $a_{n+1}\geq {\frac{1}{3}}$, so if
$$\frac{4a_n}{3a_n+3}\geq {\frac{1}{3}}$$ but this is equivalent to $$ 4a_n \geq a_n+1$$ or $$3a_n\geq 1$$ which is true by induction assumption.