<s>[INST] Find asymptotic behavior of $a_n$ We are given $a_1=1$ and $a_{n+1}=\dfrac 1 {\sqrt n \ a_n}$.
At first I found through continuous analogue that $a_n=2\sqrt{\sqrt n}$.
Then I got the first inequality : $a_{n+1}^2 -a_1 ^2 >2 \left(1+ \dots +\dfrac 1 {\sqrt n} \right)$.
Then I obtained $a_n^2 > \dfrac {2n}{\sqrt n}$.
For the second inequality I obtained $a_{n+1}^2 - a_1^2 < \dfrac 2 {\sqrt  n}+\dfrac 1 {2 \sqrt n} \left(1+\dots+\dfrac 1n\right)$,
but I am not sure if I approached the asymptotes of $a_n$ correctly.
 [/INST] It follows by induction that for any $n\geq 2$,
$$
a_{2n}  = \frac{{\sqrt {2n - 2}  \cdots \sqrt 2 }}{{\sqrt {2n - 1} \sqrt {2n - 3}  \cdots \sqrt 3 }} = \sqrt {\frac{{(2n - 2)!!}}{{(2n - 1)!!}}} 
$$
and
$$
a_{2n + 1}  = \frac{{\sqrt {2n - 1}  \cdots \sqrt 3 }}{{\sqrt {2n} \sqrt {2n - 2}  \cdots \sqrt 2 }} = \sqrt {\frac{{(2n - 1)!!}}{{(2n)!!}}} .
$$
Then, by using for example Stirling's formula,
$$
a_{2n}  \sim \sqrt {\frac{1}{2}\sqrt {\frac{\pi }{n}} } ,\quad a_{2n + 1}  \sim \frac{1}{{\sqrt {\sqrt {\pi n} } }}
$$
as $n\to +\infty$.
</s>