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: $a_1>0$
 A: 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,
$$
\bbox[5px,border:2px solid #C0A000]{a_n\ge\sqrt{n}\min\!\left(a_1,\frac1{a_1\sqrt2}\right)}
$$
A: You can prove by induction that
$$
a_{2n+1} = \frac{(2n)!!}{(2n-1)!!}\, a_1,
\qquad
a_{2n+2} = \frac{(2n+1)!!}{(2n)!!}\, \frac{1}{a_1}.
$$
Consider the subsequence $(a_{2n+1})$. You have that
$$
a_{2n+1} = a_1\, \prod_{k=1}^n \left(1 + \frac{1}{2k-1}\right)
\geq a_1\, \sum_{k=1}^n \frac{1}{2k-1}
$$
and the last sum goes to $+\infty$ as $n\to +\infty$.
An analogous computation shows that also $(a_{2n+2})$ diverges to $+\infty$.
A: We have:  $a_1a_2 = 1 $ , $a_2a_3 = 2$ ,..., $a_na_{n+1}$ = $n$.
Thus ,  $a_n = \dfrac{a_na_{n-1}}{a_{n-1}}= \dfrac{n-1}{a_{n-1}}= a_{n-2}\dfrac{n-1}{a_{n-1}a_{n-2}}=a_{n-2}\dfrac{n-1}{n-2}= \dfrac{(n-3)(n-1)}{(n-2)a_{n-3}}= a_{n-4}\dfrac{(n-1)(n-3)}{(n-2)(n-4)}= ....=\dfrac{(n-1)(n-3)\cdots 3\cdot 1}{(n-2)(n-4)\cdots 4\cdot 2}a_2= \dfrac{(n-1)(n-3)\cdots 3\cdot 1}{(n-2)(n-4)\cdots 4\cdot 2\cdot a_1} $.
Can you manage to find the limit from this point?
A: Hint. The given sequence admits a closed-form. One may observe that $a_k>0.$
From
$$
a_k\cdot a_{k+1}=k,\qquad k=1,2,3,\cdots,
$$ one gets
$$
(-1)^k\ln(a_k)-(-1)^{k+1}\ln(a_{k+1})= (-1)^k\ln k
$$ then, by summing from 1 to $n-1$, terms telescope 
$$
-\ln a_1-(-1)^n\ln(a_n)=\sum_{k=1}^{n-1}(-1)^k\ln k
$$ giving

$$
\begin{align}
a_{2n}=&\dfrac {1\cdot 3\cdots (2n-1)} {2\cdot 4\cdots (2n-2)} \cdot \frac1{a_1}=\frac{(2n)!}{2^{2n}n!^2} \cdot \frac{2n}{a_1}
\\\\
a_{2n+1}=&\dfrac {2n}{a_{2n}} =\frac{2^{2n}n!^2}{(2n)!}\cdot a_1
\end{align}
$$ 

Then one may conclude by using the Stirling formula.
A: 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.
A: Suppose the limit exists and is equal to $c$. That is $$ \lim_{n \rightarrow \infty } a_n=c.$$ Now write $a_n=\frac{n}{a_{n+1}}$. Then taking limits we get $$c=\dfrac{\lim_{n \rightarrow \infty}n}{c}$$. Hence $c$ is arbitrarily large which means $$ \lim_{n \rightarrow \infty } a_n=\infty.$$
