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 $a_n\cdot a_{n+1}=n$ and $a_{n+1}\cdot a_{n+2}=n+1$,
$$
\frac{a_{n+2}}{a_n}=\frac{n+1}n\ge\sqrt{\frac{n+2}n}
$$
Therefore, for even $n$
$$
\begin{align}
a_n
&\ge a_2\sqrt{\frac n2}\\
&=\frac1{a_1}\sqrt{\frac n2}
\end{align}
$$
and for odd $n$
$$
a_n\ge a_1\sqrt{n}
$$
Thus,
$$
\bbox[5px,border:2px solid #C0A000]{a_n\ge\sqrt{n}\min\!\left(a_1,\frac1{a_1\sqrt2}\right)}
$$
