Prove that the sequence is bounded and subsequences converge Sequence is defined in the following way:
$x_1 = \frac{2}{3}$, $x_{2n+1} = \frac{x_{2n}}{3} + \frac{2}{3} $, $x_{2n} = \frac{x_{2n-1}}{3}$
I need to show that the sequence is bounded, i.e  $0<x_{n}<1$, and it does not converge. 
This is what I have done so far:
I have separated odd and even elements as the following subsequences:
$x_{2n+1} = \frac{x_{2n-1}}{9} + \frac{2}{3} $
$x_{2n} = \frac{x_{2n-2}}{9} + \frac{2}{9} $  
And I have found that $x_{2n+1} \to \frac{3}{4} $ and  $x_{2n} \to \frac{1}{4} $
I know that $x_n$ is divergent, as the subsequential limits are not equal, but I have trouble showing that:


*

*$x_n$ is bounded from above. Do I need to look at subsequences separately and show that they are monotone or do I need to somehow estimate $x_n$?

*Showing that $x_{2n}$ and $x_{2n+1}$ are convergent


We know that $$x_1=\dfrac{2}{3}\\x_{2n+1}=\dfrac{x_{2n}}{3}+\dfrac{2}{3}\qquad\qquad (I)\\x_{2n-1}=\dfrac{x_{2n-2}}{3}+\dfrac{2}{3}\qquad\qquad (II)\\x_{2n}=\dfrac{x_{2n-1}}{3}\qquad\qquad (III)$$by substituting $(II)$ in $(III)$ and $(III)$ in $(I)$ we respectively obtain$$x_{2n}=\dfrac{x_{2n-2}+2}{9}\\x_{2n+1}=\dfrac{x_{2n-1}+6}{9}$$which yields to $x_{2n}\to \dfrac{1}{4}$ and $x_{2n+1}\to \dfrac{3}{4}$ using $\epsilon-\delta$ approach (check it!) and $x_n$ is divergent.
For showing boundedness using induction, note that $0<x_1<1$. Also if $0<x_{2n}<1$ then $x_{2n+1}=0<\dfrac{2}{3}<\dfrac{x_{2n}+2} {3}<1$ and if $0<x_{2n+1}<1$ then $0<x_{2n+2}=0<\dfrac{x_{2n+1}}{3}<\dfrac{1}{3}<1$ which yields to $0<x_n<1$.
