Define a sequence $(a_n)$ as follows $a_1 = 1$ and for $n\geq 1,\;a_{n+1} = (-1)^n\frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)$ Define  a sequence $(a_n)$ as follows $a_1 = 1$ and for $n\geq 1$ 
$$a_{n+1} = (-1)^n\frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)$$
then what will be it's $\limsup (a_n)$ and $\liminf (a_n)$ and $\sup(a_n)$, $\inf(a_n)$.
I tried to determine the nature of $a_{2n+1}$ and $a_{2n}$. I got $a_{2n+1}$ is increasing and $a_{2n+1} \geq \text{something}$ where $a_{2n}$ is decreasing  and $a_{2n}\leq \text{something}$. So I could not draw any conclusion. 
Can anyone please help me?

I think the following will help the readers to understand the solution:
We know  $$\frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)\ge\sqrt{|a_n| \times \frac {2}{|a_n|}}=\sqrt 2$$
$$\implies \frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)\ge\sqrt 2$$
Given that $~a_1 = 1~$ and for $n\geq 1$
$$a_{n+1} = (-1)^n\frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)$$
When $~n~$ is even, $$a_{n+1} = \frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)\ge\sqrt 2$$
Again when $~n~$ is odd, $$a_{n+1} = -\frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)\le-\sqrt 2$$
Which gives $~\lim\sup a_n=\sqrt 2~$ and $~\lim\inf a_n=-\sqrt 2~$.
