Proof that non-monotonic sequence is unbounded I am looking at a non-monotonic recursive sequence given by
$$a_0 =2,$$
$$a_{n+1}  = \frac{1}{2}\left( a_n - \frac{1}{a_n}\right).$$
Is the sequence bounded or unbounded?
I suspect that the sequence is unbounded.
Easy concepts like finding another unbounded sequence as a lower bound fails. This is because the sequence always approaches 0 until $a_n$ gets very small at which point the sequence blows up again. Any ideas how to go from here?
Note: I saw this problem in some public talk for high school students around 12 years ago, and it got stuck in my head ever since (the speaker even set a reward for solving this). Maybe this is some kind of trick question that relates to some famous unanswered problem so there will be no hope for mere mortals like me. Any hints in that direction would be nice as well.
Hint:
Establish that liminf an = -∞ as n-&gt;∞ and
            limsup an = +∞ as n-&gt;∞.

knowing these two facts, what can we say about the boundedness of sequence $a_n$
The sequence $a_n$ is unbounded,  -∞ &lt; $a_n$ &lt; ∞.
and liminf $a_n$ = -∞ as n-&gt;∞ and limsup $a_n$ = +∞ as n-&gt;∞.
Proof
The solution for $a_n$ is   $a_n$ =  cot(arccot(2)$2^n$)
or                          $a_n$          ~  cot(.4636476*$2^n$)
consider the the function a(x) in interval (nπ, (n + 1)π) as x-&gt;∞.
Define a(x) as a(x) := cot ((arccot(2)2x)), x ε R,
then a(x)-&gt; -∞ as x-&gt;(nπ)+, and a(x)-&gt; ∞, as x-&gt; ((n + 1)π)-.
The process is repeated in every interval (n π, (n+ 1)π) of duration π, n = 0, 1, 2, 3,...
It follows that a_n -&gt; -∞ infinitely often and a_n -&gt; +∞ infinitely often.
Therefore
liminf $a_n$ = -∞ as n-&gt;∞, and limsup $a_n$ = +∞ as n-&gt;∞ as asserted.
Graph of $a_n$ for n= -1023 to 1063