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 !
 A: We find $a_1/a_3=1/2$, $a_1a_4=1\cdot3/2$, $a_1/a_5=1\cdot3/(2\cdot4)$.
By induction, it one shows that
$$a_1a_{2n}=\frac{1\cdot3\cdot\dots\cdot(2n-1)}{2\cdot4\cdot\dots\cdot(2n-2)}$$
for integer $n\geq2$, whereas
$$a_1/a_{2n+1}=\frac{1\cdot3\cdot\dots\cdot(2n-1)}{2\cdot4\cdot\dots\cdot(2n)}$$
for integers $n\geq1$. 
Since $a_{2n}/a_{2n+1}\to1$ by assumption, we have
$$a_1^2=\lim_{n\to\infty} \frac{(1\cdot3\cdot\dots\cdot(2n-1))^2}
    {(2\cdot4\cdot\dots\cdot(2n-2))^2(2n)}=
    \lim_{n\to\infty}\frac{(2n-1)!^2}{2^{4n-3}(n-1)!^4n}=
    \lim_{n\to\infty}\frac{(2n)!^2n}{2^{4n-1}n!^4}.$$
A calculation using Stirling's formula shows that
$a_1^2=2/\pi$ and hence 
$$|a_1|=\sqrt{\frac{2}{\pi}}.$$
Edit: a) Let me complete the solution by a calculation of the limit using Stirling's formula twice:
$$\begin{array}{rcl}2\log((2n)!)-4\log(n!)-(4n-1)\log2+\log n&
=&2\left[(2n+\frac12)(\log n+\log2)-2n+\frac12\log(2\pi)\right]\\
&&-4\left[(n+\frac12)\log n-n+\frac12\log(2\pi)\right]\\
&&+\log n-(4n-1)\log2+ O(\frac1n)\\
&=&2\log 2-\log(2\pi)+O(\frac1n)\to\log(2/\pi).\end{array}$$
b) The solution also works in the complex domain. Then we obtain $a_1=\pm\sqrt{\frac2\pi}$ which is more precise in the complex domain.
c) For arbitrary $a_1$, the solution shows that
$$\frac{a_{2n}}{a_{2n+1}}\to\frac2{\pi a_1^2}.$$
As $\frac{a_{2n}}{a_{2n+2}}=\frac{2n}{2n+1}\to1$, this implies that
$\frac{a_{2n+1}}{a_{2n+2}}\to\frac{\pi a_1^2}2.$
Therefore $a_n/a_{n+1}\to1$ if and only if $a_1^2=2/\pi$.
d) One can also write using the first two formulas of the solution and $a_{2n+2}/a_{2n+1}\to1$
$$a_1^2=\lim_{n\to\infty}a_1^2\frac{a_{2n+2}}{a_{2n+1}}=\lim_{n\to\infty}\frac{(1\cdot3\cdot\dots\cdot(2n-1))^2(2n+1)}
    {(2\cdot4\cdot\dots\cdot(2n-2)(2n))^2}=\frac{1\cdot3}{2^2}\cdot
    \frac{3\cdot5}{4^2}\cdots=\frac2\pi$$
by the Wallis product.
A: 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 $a_1= \sqrt{2/\pi}$.
A: You may notice that a celebrated sequence has very similar properties: by defining
$$ I_n = \int_{0}^{\pi/2}\left(\sin \theta\right)^n\,d\theta $$
we have
$$ I_{2n} = \frac{\pi}{2\cdot 4^n}\binom{2n}{n},\qquad I_{2n+1}=\frac{4^n}{(2n+1)\binom{2n}{n}} $$
by integration by parts. $\{I_n\}_{n\geq 1}$ is obviously decreasing to zero and fulfills $ I_n I_{n+1} = \frac{\pi}{2(n+1)}$. Moreover, by Euler's Beta function (or just from the previous line) we have
$$ I_n = \frac{\sqrt{\pi}\,\Gamma\left(\frac{n+1}{2}\right)}{2\,\Gamma\left(\frac{n+2}{2}\right)}\sim\sqrt{\frac{\pi}{2n}}. $$
It follows that the sequence
$$ a_n = \frac{\sqrt{\pi/2}}{\int_{0}^{\pi/2}\left(\sin\theta\right)^{n-1}\,d\theta} $$
meets the hypothesis of our problem. Provided that the solution is unique, this proves $a_1=\sqrt{\frac{2}{\pi}}$ without invoking Stirling's approximation.
