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.
 A: EDIT: This response was written for the recurrence equations
$$ a_{2n} = \frac{a_{2n-1}}{2}, a_{2n+1} = \frac{1}{2}+\frac{a_{2n}}{2}. $$
The exact answer may differ depending on the exact edit to the question, but the techniques remain valid, assuming the coefficients are not drastically changed.

Since there are different rules for the even and odd terms, it makes sense to suspect that the even terms will behave differently from the odd terms. So let's investigate the two sets of terms separately.
For $n\in\mathbb{N}$, let $b_n = a_{2n-1}$ and $c_n = a_{2n}$. From the relations provided, we have
$$ b_{n+1} = a_{2n+1} = \frac{1}{2}+\frac{a_{2n}}{2} = \frac{1}{2} + \frac{1}{2}\left(\frac{a_{2n-1}}{2}\right) = \frac{1}{2} + \frac{b_{n}}{4}$$
Similarly, we have
$$ c_{n+1} = a_{2n+2} = \frac{a_{2n+1}}{2} = \frac{1}{2}\left(\frac{1}{2} + \frac{a_{2n}}{2}\right) = \frac{1}{4} + \frac{c_n}{4}.$$
We now look at the behavior of $b_n$ as $n\rightarrow\infty$. Suppose it converged to some value, say $b$. What would $b$ have to be? If we take our equation
$$ b_{n+1} = \frac{1}{2} + \frac{b_n}{4} $$
and let $n\rightarrow\infty$, then both $b_{n+1}$ and $b_n$ would converge to $b$, and hence
$$ b = \frac{1}{2} + \frac{b}{4}\implies b=\frac{2}{3}.$$
So we have a hunch that $b_n$ converges to $\frac{2}{3}$. How do we prove it? Well, let $d_n = b_n - \frac{2}{3}$, and let's try to show that $d_n\rightarrow 0$. Now,
$$ b_{n+1} = \frac{1}{2} + \frac{b_n}{4}\implies d_n + \frac{2}{3} = \frac{1}{2} + \frac{d_n + \frac{2}{3}}{4} = \frac{2}{3} + \frac{d_n}{4}\implies d_{n+1} = \frac{d_n}{4}$$
which clearly shows that $d_n\rightarrow 0$. Hence, $b_n\rightarrow\frac{2}{3}$. Similar reasoning yields that $c_n\rightarrow c$, where
$$ c = \frac{1}{4} +\frac{c}{4}\implies c = \frac{1}{3}.$$
Hence, the odd terms tend to $\frac{2}{3}$, and the even terms tend to $\frac{1}{3}$. This shows that
$$ \limsup\limits_{n\rightarrow\infty}{a_n} = \frac{2}{3} $$
and
$$ \liminf\limits_{n\rightarrow\infty}{a_n} = \frac{1}{3}. $$
A: It is not possible to get the limsup from a recursive formula, which is how it has been given in the problem statement. However, through an inductive proof, an explicit formula for the terms of this sequence can be found: 
$a_{2n} = \frac {1}{2}  - \frac{1}{2^n}$
$a_{2n+1} = 1 - \frac{1}{2^n}$
Taking the limits of both of these subsequences gives $\frac{1}{2}$ and 1, respectively.
The limsup $a_n$ is ≥ 1. (Since we have a subsequence converging to 1, and the limsup is the biggest subsequential limit, it can not be less than 1.)  Further, the only two subsequential limits we can get are  1 and 1/2. So the limsup is 1.
This is because these two subsequences together contain all of the terms of $a_n$, (that is, every term in the sequence $a_n$  is either $a_{2n} $ and $a_{2n+1}$ , since n is either even or odd).
An alternative way to think of this is that the set of subsequential limits is the set {$1 , \frac{1}{2}$}. Every convergent subsequence converges to one of these two values. Then, the supremum of this set is 1, so limsup $a_n$ is 1.
A: 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.
