If $a_1a_2 = 1, a_2a_3 = 2, a_3a_4 = 3 \cdots$ and $ \lim_{n \to \infty} \frac{a_n}{a_{n+1}} = 1$, find $|a_1|$ 
If $a_1a_2 = 1, a_2a_3 = 2, a_3a_4 = 3 \cdots$ and $\displaystyle \lim_{n \to \infty} \frac{a_n}{a_{n+1}} = 1$. Find $|a_1|$

I could conclude that $\displaystyle \lim_{n \to \infty}a_n$ must be $\infty$. But couldn't get any further. Any help is welcome.
 A: Note that
$$
\frac{a_3}{a_1} = \frac{a_2a_3}{a_1a_2} = \frac21,\quad
\frac{a_5}{a_3} = \frac{a_4a_5}{a_3a_4} = \frac43,\quad
\frac{a_7}{a_5} = \frac{a_6a_7}{a_5a_6} = \frac65,\quad\dots
$$
and therefore
$$
\frac{a_{2k+1}}{a_1} = \frac{a_3}{a_1} \frac{a_5}{a_3}\cdots \frac{a_{2k+1}}{a_{2k-1}} = \frac21 \frac43 \cdots \frac{2k}{2k-1} = \frac{4^k(k!)^2}{(2k)!}.
$$
Similarly,
$$
\frac{a_{2k+2}}{a_2} = \frac{a_4}{a_2} \frac{a_6}{a_4} \cdots \frac{a_{2k+2}}{a_{2k}} = \frac32 \frac54 \cdots \frac{2k+1}{2k} = \frac{(2k+1)!}{4^k(k!)^2}.
$$
Therefore
$$
1 = \lim_{k\to\infty} \frac{a_{2k+1}}{a_{2k+2}} = \lim_{k\to\infty} \frac{4^k(k!)^2 a_1/(2k)!}{(2k+1)!a_2/4^k(k!)^2} = \lim_{k\to\infty} \frac{a_1}{a_2} \frac{16^k(k!)^4}{(2k)!(2k+1)!} = \frac{a_1}{a_2} \frac\pi2
$$
(using Stirling's formula),
which forces $a_1/a_2 = 2/\pi$; together with $a_1a_2=1$ this yields $a_1=\pm\sqrt{2/\pi}$. (One can reality check $\lim_{k\to\infty} \frac{a_{2k}}{a_{2k+1}}$ as well if desired.)
A: By the Wallis product
\begin{align*}
\frac{2}{\pi } & = \mathop {\lim }\limits_{N \to  + \infty } \prod\limits_{n = 1}^N {\frac{{(2n - 1) \cdot (2n + 1)}}{{2n \cdot 2n}}}  = \mathop {\lim }\limits_{N \to  + \infty } \prod\limits_{n = 1}^N {\frac{{a_{2n - 1} a_{2n} a_{2n + 1} a_{2n + 2} }}{{a_{2n} a_{2n + 1} a_{2n} a_{2n + 1} }}} \\ & = \mathop {\lim }\limits_{N \to  + \infty } \left( {\prod\limits_{n = 1}^N {\frac{{a_{2n - 1} }}{{a_{2n} }}} \prod\limits_{n = 1}^N {\frac{{a_{2n + 2} }}{{a_{2n + 1} }}} } \right) = \mathop {\lim }\limits_{N \to  + \infty } \left( {\prod\limits_{n = 1}^N {\frac{{a_{2n - 1} }}{{a_{2n} }}} \prod\limits_{n = 2}^{N + 1} {\frac{{a_{2n} }}{{a_{2n - 1} }}} } \right) \\ & = \frac{{a_1 }}{{a_2 }}\mathop {\lim }\limits_{N \to  + \infty } \frac{{a_{2N + 2} }}{{a_{2N + 1} }} = \frac{{a_1 }}{{a_2 }}.
\end{align*}
Thus,
$$
\frac{2}{\pi } = \frac{{a_1 }}{{a_2 }} = \frac{{a_1^2 }}{{a_1 a_2 }} = a_1^2 ,
$$
i.e.,
$$
\left| {a_1 } \right| = \sqrt {\frac{2}{\pi }} .
$$
