Finding the limit of the sequence $a_n\cdot a_{n+1}=n,\,n=1,2,3,\cdots.$ Let $(a_n)_{n>=1}$ be a sequence of real numbers defined by the below recurrence relation:

$$a_n\cdot a_{n+1}=n,\quad n=1,2,3,\cdots.$$

Prove that $\lim_{n\to \infty}a_n=+\infty.$
Edit: $a_1>0$

Since $a_1 > 0$, it follows that $a_n > 0$, for all $n$.

Since $a_n a_{n+1} = n$, at least one of $a_n,a_{n+1}$ is greater or equal to $\sqrt{n}$, hence the sequence $(a_n)$ is unbounded above.

For all positive integers $n > 2$, we have
\begin{align*}
a_n &= \frac{n-1}{a_{n-1}}\\[4pt]
a_{n-2} &= \frac{n-2}{a_{n-1}}\\[4pt]
\end{align*}
hence
$$
\frac
{a_n}
{a_{n-2}}
=
\frac
{n-1}{n-2}
$$
It follows that
$$a_1 < a_3 < a_5 < \cdots$$
$$a_2 < a_4 < a_6 < \cdots$$
Hence, since the sequence $(a_n)$ is unbounded above, at least one of the above subsequences approaches infinity.

But for $m>1$, we have
\begin{align*}
a_{2m+2} &\,=\, 
a_2\prod_{k=1}^m \frac{a_{2k+2}}{a_{2k}}
\,=\,
a_2\prod_{k=1}^m \frac{2k+1}{2k}
\\[4pt] 
a_{2m+1} &\,=\, 
a_1\prod_{k=1}^m \frac{a_{2k+1}}{a_{2k-1}}
\,=\,
a_1\prod_{k=1}^m \frac{2k}{2k-1}\\[4pt] 
\end{align*}
so
$$
\frac
{a_{2m+1}}{a_{2m+2}} 
= \frac{a_1}{a_2}
\prod_{k=1}^m \frac{4k^2}{4k^2-1} > \frac{a_1}{a_2}
$$
Hence, since at least one of the every-other-term subsequences approaches infinity, it follows that the odd-term subsequence approaches infinity.

But also,
$$
a_{2m+3} \,=\, 
a_3\prod_{k=1}^{m} \frac{a_{2k+3}}{a_{2k+1}}
\,=\, a_3\prod_{k=1}^{m} \frac{2k+2}{2k+1}
$$
so
$$
\frac
{a_{2m+2}}{a_{2m+3}} 
= \frac{a_2}{a_3}
\prod_{k=1}^m \frac{4k^2+4k+1}{4k^2+4k} > \frac{a_2}{a_3}
$$
Hence, since the odd-term subsequence approaches infinity, the even-term subsequence must also approach infinity.

It follows that the sequence $(a_n)$ approaches infinity.
