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.

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.,
$$
\left| {a_1 } \right| = \sqrt {\frac{2}{\pi }} .
$$
