The sequence is given by: $$a_{1} = 3/2 , a_{n+1} = \sqrt {3a_{n} - 2}$$
1-I tried to find the limit and calculations lead me to either the limit $L = 2$ or $L =1$, so then how can I choose between them?
2- For studying monotonicity of this sequence I got:
$$a_{n+1} - a_{n} = \frac{3(a_{n} - a_{n-1})}{a_{n+1} + a_{n}} $$
Then how can I know that the value is greater than or less than zero?
3-Finally how can I study boundedness?
On
HINT
We have that for
$$1<a_n<2 \implies 1<\sqrt {3a_{n} - 2}<2$$
therefore the sequence is bounded, moreover we have
$$a_{n+1}-a_n = \sqrt {3a_{n} - 2}-a_n=(\sqrt {3a_{n} - 2}-a_n)\frac{\sqrt {3a_{n} - 2}+a_n}{\sqrt {3a_{n} - 2}+a_n}=$$$$=\frac{-a_n^2+3a_n-2}{\sqrt {3a_{n} - 2}+a_n}=\frac{-(a_n-1)(a_n-2)}{\sqrt {3a_{n} - 2}+a_n}>0$$
Observe that $a_1<a_2$. In general, show that $a_{n-1}<a_n$ implies $\sqrt{3a_{n-1}-2}<\sqrt{3a_n-2}$ etc.
Also $a_1<2$. Show that $a_n<2$ implies $\sqrt{3a_n-2}<2$ etc.