<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] 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.
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