Given $a_1 \in \mathbb R$ . Consider the sequence $\{a_n\}$, which is recursively defined by$$ a_{n+1} =\left\{\begin{array}{ccc} \dfrac {a_n}{2} & \text{if $n$ is even,}\\ \dfrac {1 + a_n }{2} & \text{if $n$ is odd}.\end{array} \right. $$
Find the sub-sequential limit of this sequence.
I think for odd $n$ the limit point will be $l= \frac{1+l}{2}$ now $2l = 1+l$ , here $ l = 1$
I don't know the limit point for0 even $n$.
Please help me.
Observe that $$ a_{n+2}=\left\{\begin{array}{lll} \dfrac{a_n+1}{4} &\text{if $n$ odd} \\ \dfrac{a_n+2}{4} &\text{if $n$ even} \end{array} \right. $$ It can be readily shown that each of the subsequence, $\{a_{2n}\}$ and $\{a_{2n-1}\}$ is monotonic and bounded, and hence convergent.
Hence $$ a_{2n}\to \frac{2}{3}, \quad\text{while}\quad a_{2n-1}\to \frac{1}{3}. $$