<s>[INST] On infinite sequences that satisfy $a_na_{n+1}=n$ and $\liminf _{n\rightarrow \infty}{\frac{a_{n+1}}{a_{n}}}=1$ 
Question: 
   Determine all infinite convergent
  sequence $\{a_{n}\}_{n\ge 1}$, such that the following two conditions are satisfied.
  
  
*
  
*$a_{n}a_{n+1}=n$
  
*$\displaystyle\liminf_{n\to\infty}\dfrac{a_{n+1}}{a_{n}}=1$
  

This problem is from china analysis 600 problem book page 488.
My idea: since
$$a_{n}a_{n+1}=n$$
$$a_{n+1}a_{n+2}=n+1$$
so
$$\dfrac{a_{n+2}}{a_{n}}=\dfrac{n+1}{n}$$
Case one: if $n$ is even postive integer,then
$$\dfrac{a_{n}}{a_{2}}=\dfrac{1\cdot 3\cdot 5\cdots (n+1)}{2\cdot 4\cdot 6\cdots n}=\dfrac{(n+1)!!}{n!!}$$
Case 2: if $n$ is odd postive integer,then
$$\dfrac{a_{n}}{a_{1}}=\dfrac{2\cdot 4\cdot 6\cdots(n+1)}{1\cdot 3\cdot 5\cdots n}=\dfrac{(n+1)!!}{n!!}$$
then I can't how to use the second condition
 [/INST] Actually
\begin{align*}&a_{2k+1}=\frac{2k}{2k-1}a_{2k-1}=\ldots=\frac{(2k)!!}{(2k-1)!!}a_1\\
&a_{2k+2}=\frac{2k+1}{2k}a_{2k}=\ldots=\frac{(2k+1)!!}{(2k)!!}a_2\end{align*}
Observe that (see Wallis product)
$$\lim_{k\to\infty}\frac{a_{2k+2}}{a_{2k+1}}=\frac{a_2}{a_1}\lim_{k\to\infty}\frac{(2k+1)!!(2k-1)!!}{((2k)!!)^2}=\frac{a_2}{a_1}\prod_{k=1}^\infty\frac{(2k+1)(2k-1)}{(2k)^2}=\frac{a_2}{a_1}\cdot\frac2\pi$$
Then
$$\lim_{k\to\infty}\frac{a_{2k+1}}{a_{2k}}=\lim_{k\to\infty}\frac{a_{2k+2}}{a_{2k}}\cdot\frac{a_{2k+1}}{a_{2k+2}}=\lim_{k\to\infty}\tfrac{2k+1}{2k}/\lim_{k\to\infty}\frac{a_{2k+2}}{a_{2k+1}}=\frac{a_1}{a_2}\cdot\frac\pi2$$
So the second condition implies
$$\frac{a_2}{a_1}\ge\frac\pi2\text{ and }\frac{a_1}{a_2}\ge\frac2\pi,$$
hence $\frac\pi2\le\frac{a_2}{a_1}\le\frac\pi2$ and also $a_1a_2=1$, so all sequences satisfying (1) and (2) are given by $a_1=\pm\sqrt{\frac2\pi}$ and $a_2=\pm\sqrt{\frac\pi2}$.
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