How do I find $\liminf$ and $\limsup$ if $a_{2n}=\frac {a_{2n-1}}2$ and $a_{2n+1}=\frac12+\frac {a_{2n}}2$? Its given that $a_1=a>0$ and that for any $n>1$ two things happen:
$$a_{2n}=\frac {a_{2n-1}}2$$
$$a_{2n+1}=\frac12+\frac {a_{2n}}2$$
How do I find $\lim\inf$ and $\lim\sup$
I am trying to look at 
$a_{2n+1}$ and $a_{2n-1}$
$a_{2n}$ and $a_{2n-2}$
But I am unable to prove that they are bounded.
NOTE: Look at joeys answer for correct solution.

It could of course be that I made a mistake in my calculations so proceed with caution!
We'll show by induction that for any $n \geq 1$:
$$
a_{2n} =  \frac{a}{2\cdot 4^n} + \sum_{i=1}^{n-1} \frac{1}{4^i}
$$
where a sum $\sum_{i=1}^0\frac{1}{4^n}$ is assumed to be zero.
The first step: $n=1$ is trivial. Now let $n$ be arbitrary we'll first prove the form of $a_{2(n+1)}$: by the induction hypethesis we may assume that:
$
a_{2n} =\frac{a}{2\cdot 4^n} + \sum_{i=1}^{n-1} \frac{1}{4^i}
$ thus we get:
$$
a_{2(n+1)} = a_{2n+2} = \frac{a_{2n+1}}{2} = \frac{\frac{1}{2} + \frac{a_{2n}}{2}}{2} = \frac{1}{4} + \frac{a_{2n}}{4} \overset{IH}{=}  \frac{1}{4} + \frac{\frac{a}{2\cdot 4^n} + \sum_{i=1}^{n-1} \frac{1}{4^i}}{4} \\= \frac{1}{4} + \frac{a}{2\cdot 4^{n+1}} + \sum_{i=1}^{n-1} \frac{1}{4^{i+1}}= \frac{a}{2\cdot 4^{n+1}} + \sum_{i=1}^{n} \frac{1}{4^{i}}
$$
which proves the claimed equality for $a_{2n}$ we now have for arbitrary $n$:
$$
a_{2n+1} = \frac{1}{2} + \frac{a_{2n}}{2} = \frac{1}{2} + \frac{\frac{a}{2\cdot 4^n} + \sum_{i=1}^{n-1} \frac{1}{4^n}}{2} = \frac{a}{4^{n+1}} + \frac{1}{2} \sum_{i=0}^n \frac{1}{4^i}.$$
Can you deduce the result from this? I can work it out further if you want but I think it's clear how you should continue now (just calculating the sums yields that the liminf is 1/3 and the limsup is 2/3).
Now without liminf and limsup, first we show that $a_n \geq 1/3 $ always holds by induction. For $a_{2n+1}$ this is immediately clear since we have $a_{2n+1} \geq 1/2$. For $a_{2n}$ we proceed by induction so suppose we have $a_{2n} \geq 1/3$ and we show that then also $a_{2(n+1)} \geq 1/3$, we have:
$$
a_{2(n+1)} = a_{2n + 2} = \frac{a_{2n-1}}{2} = \frac{1}{4} + \frac{a_{2n}}{4} \geq \frac{1}{4} + \frac{\frac{1}{3}}{4} = 1/3,
$$
in the same way one shows that $a_n \leq 2/3$ for all $n$. Now it remains to show that $\lim_{n \rightarrow \infty} a_{2n} \leq 1/3$ and $\lim_{n\rightarrow \infty} a_{2n+1} \geq 2/3$ if you show this you are done.
