<s>[INST] 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.
 [/INST] 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 $L=-l=\sqrt 2.$
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