limit superior for given recurrence for sequence Sequence ${a_n}$ is defined as follows: 
$a_1 = 1$ and for $n \geq 1,\, a_{n+1} = (-1)^n \left( \frac {1}{2} \right) \left(|a_n|+ \frac{2}{|a_n|}\right).$ Then find $\lim \sup a_n$ and $\lim \inf a_n .$
How to consider subsequences here? And how to obtain a simplified recurrence? The sequence has alternate positive and negative terms. Please help.
 A: An induction shows that  $|a_{n+1}| = \left( \frac {1}{2} \right) \left(|a_n|+ \frac{2}{|a_n|}\right)\le \left( \frac {1}{2} \right) \left(2+ \frac{2}{1}\right)=2$ and $|a_{n+1}| = \left( \frac {1}{2} \right) \left(|a_n|+ \frac{2}{|a_n|}\right)\ge \left( \frac {1}{2} \right) \left(1+ \frac{2}{2}\right)=1$ so $1\le |a_n|\le 2.$ 
Thus, $\limsup a_n=L$ and $\liminf a_n=l$ are real numbers.
Now, $(a_n)$ is alternating, the even terms are negative and the odd terms are positive. Therefore, 
$L=\limsup a_n=\limsup a_{2n+1}=\limsup \left(\frac{|a_{2n}|}{2}+\frac{1}{|a_{2n}|}\right )=$
$\limsup \left(\frac{-a_{2n}}{2}+\frac{1}{-a_{2n}}\right )=-\liminf \left(\frac{a_{2n}}{2}+\frac{1}{a_{2n}}\right ).$ 
But $\liminf a_n=\liminf a_{2n}$ so the above becomes $-L=\frac{l}{2}+\frac{1}{l}.$ Similarly, $-l=\frac{L}{2}+\frac{1}{L}.$ Solving these equations gives $L=-l=\sqrt 2.$
A: First note that if $a_n\neq0$, then $|a_n|+\frac{2}{|a_n|}>0$, so $a_{n+1}\neq0$. Considering $a_1=1$, by induction, $a_n\neq0$ for all $n$.
We have $a_1=1,\;a_2=-\frac32,\;a_3=\frac{17}{12}$. If we assume that $1\leq|a_n|\leq2$, then, inverting,
$$\frac11\geq\frac{1}{|a_n|}\geq\frac12$$
doubling,
$$2\geq\frac{2}{|a_n|}\geq1$$
reversing,
$$1\leq\frac{2}{|a_n|}\leq2$$
averaging with the original,
$$\frac{1+1}{2}\leq\frac12\bigg(|a_n|+\frac{2}{|a_n|}\bigg)\leq\frac{2+2}{2}$$
$$1\leq|a_{n+1}|\leq2$$
so by induction, all $|a_n|$ are in the interval $[1,2]$.
If we assume that $|a_n|$ has a limit $L$, then
$$L=\lim |a_n|=\lim |a_{n+1}|=\lim\frac12\bigg(|a_n|+\frac{2}{|a_n|}\bigg)=\frac12\bigg(L+\frac{2}{L}\bigg)$$
which is a quadratic equation giving $L=\sqrt2$ . Indeed, we can prove that it does have a limit. If $|a_n|=\sqrt2+\varepsilon_n$ (allowing $\varepsilon_n$ to be positive or negative), then
$$\varepsilon_{n+1}=|a_{n+1}|-\sqrt2$$
$$=\frac12\bigg(\frac{|a_n|^2+2}{|a_n|}\bigg)-\sqrt2$$
$$=\frac12\bigg(\frac{(2+2\sqrt2\,\varepsilon_n+\varepsilon_n^2)+2}{\sqrt2+\varepsilon_n}\bigg)-\sqrt2$$
$$=\frac{4+2\sqrt2\,\varepsilon_n+\varepsilon_n^2}{2(\sqrt2+\varepsilon_n)}-\frac{2+\sqrt2\,\varepsilon_n}{\sqrt2+\varepsilon_n}$$
$$=\frac{\varepsilon_n^2}{2(\sqrt2+\varepsilon_n)}$$
which is necessarily positive. (Thus only $\varepsilon_1$ is negative.) The denominator is between $2$ and $4$, because $|a_n|$ is between $1$ and $2$. So
$$\frac14\varepsilon_n^2\leq\varepsilon_{n+1}\leq\frac12\varepsilon_n^2$$
and if $\varepsilon_n<2$ (which we know is true for small $n$), then
$$\varepsilon_{n+1}\leq\Big(\frac12\varepsilon_n\Big)\varepsilon_n<(1)\varepsilon_n.$$
This shows that $\varepsilon_n$, and thus $|a_n|$, is decreasing. It's also bounded, so it has a limit.
Finally, using $a_n=(-1)^{n-1}|a_n|$, we get
$$\lim\sup a_n=-\lim\inf a_n=\sqrt2.$$
