<s>[INST] 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.
 [/INST] 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.)
</s>