<s>[INST] Sequence problem : Find $|a_1|$ 
Given a sequence such that
$$a_1a_2=1\ ; \  a_2a_3=2 \  ; \  a_3a_4=3  \ ; \ \dots$$ and
$$\lim_{n\to \infty} \frac{a_n}{a_{n+1}}=1$$
find $|a_1|$.

My attempts :
We can deduce from this $a_1a_2=1 \  ; \  a_2a_3=2 \  ; \  a_3a_4=3  \ ; \ \dots$ that :
$$\prod_{n=k}^{n+1} a_n=n$$
Thus :
$$\begin{align}a_n \times a_{n+1}&=n \\ a_{n+1}&=\frac{n}{a_n} \end{align}$$
Therefore :
$$\begin{align} \lim_{n\to \infty} \frac{a_n}{a_{n+1}}&=\lim_{n\to \infty} \frac{a_n}{\frac{n}{a_n}} \\ &=\lim_{n\to \infty} \frac{a_n^2}{n} \\&=1 \end{align} $$
Is this can lead us to say that :
$$\begin{align} |a_n^2|&\sim n \\ |a_n|&\sim \sqrt{n} \end{align} $$
If this is true we can say that :
$$|a_n| =1$$
But I don't know if this is true or not. Any tips or hints ?
Thanks in advance !
 [/INST] Since $a_{n+1} = n/a_n$ the sequence $a_n$ is fully determined from the starting value $a_1$. 
In fact it turns out that

 $$a_n = \frac{n-1}{a_{n-1}} =\frac{n-1}{n-2} a_{n-2} = \cdots = \frac{(n-1)!!}{(n-2)!!} a_1^{(-1)^{n+1}}$$ 

Where $N!!$ denotes the semifactorial. From this you get that

 $$\frac{a_{n}}{a_{n+1}} = \frac{((n-1)!!)^2}{(n-2)!! n !!} (a_1)^{2(-1)^{n+1}}$$ 

In order to analyze asymptotic behaviour you may want distinguish even and odd $n$
$$\frac{a_{2n}}{a_{2n+1}} = \frac{((2n-1)!!)^2}{(2n)!!(2n-2)!!} a_1^{-2} = \left(\frac{(2n-1)!!}{(2n)!!}\right)^2 \frac{2n}{a_1^2} = \left(\frac{(2n)!}{(2n )!!^2}\right)^2\frac{2n}{a_1^2}$$
$$\frac{a_{2n+1}}{a_{2n+2}} = \frac{((2n)!!)^2}{(2n+1)!!(2n-1)!!} a_1^{2} = \left(\frac{(2n)!!}{(2n-1)!!}\right)^2 \frac{a_1^2}{2n+1} = \left(\frac{(2 n)!!^2}{(2n)!}\right)^2\frac{a_1^2}{2n+1}$$
Passing to logarithms you get
$$\log\left(\frac{a_{2n}}{a_{2n+1}}\right) = 2 \big( \log( (2n)! ) - 2\log( (2n)!!) \big) +\log(2n) - 2\log(a_1)$$ 
$$\log\left(\frac{a_{2n+1}}{a_{2n}}\right) = 2 \big( 2\log( (2n)!!) - \log((2n)!)\big) - \log(2n+1) + 2\log(a_1)$$ 
Recall the Stirling approximation
$$\log(n!)=n\log(n)-n+\frac{1}{2} \log(n) + \log(\sqrt{2\pi}) +o(1)$$
from this you get approximations 
$$\log((2n)!!) = \log(2^n n!) = n\log(n) + (\log(2)-1) n + \frac{1}{2} \log(n) + \log(\sqrt{2\pi}) +o(1) $$
$$\log((2n)!) = 2n \log(n) + (2\log(2)-2) n + \frac{1}{2} \log(n) + \log(2\sqrt{\pi})+o(1)$$
Subtracting you have $2\log((2n)!!) - \log((2n)!) = \frac{1}{2} \log(n) + \log( \sqrt{\pi}) +o(1)$, so
$$\log\left(\frac{a_{2n}}{a_{2n+1}}\right) = \log(2) - \log(\pi) -2 \log(a_1) +o(1)$$
$$\log\left(\frac{a_{2n+1}}{a_{2n}}\right) = \log(\pi) - \log(2) + 2\log(a_1) +o(1)$$ 
Hence there is a value of $a_1$ for which $a_n/a_{n+1} \to 1$ and it has to satisfy $2\log(a_1)= \log(2/\pi)$, that is $a_1= \sqrt{2/\pi}$.
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