$$1\cdot \frac{1}{2}\cdot 3\cdot \frac{1}{4}\cdot 5\cdot \frac{1}{6}\cdots$$ What does this product come out to? It does diverge, but products like this tend to have values $\lt \infty$.
Here is what I would do, but I don't know if this is right.
Lets say: $$f(n)=n\cdot \frac{1}{n+1}\cdot (n+2)\cdot \frac{1}{n+3}\cdots$$ and of course in this case it must be $f(1)$ right?
Next, $$\frac{(n+1)f(n)}{n}=f(n+2)$$ Now for all $n$ replace it with $(n-2)$
(Notice that you must replace on both sides)
$$\frac{(n-1)f(n-2)}{n-2}=f(n)$$
And plug in $n=1$ back into the equation
$$\frac{(0)f(-1)}{-1}=f(1)$$ but that kinda backfires because you end up with a denominator of $0$ which is not allowed. How do you calculate this infinite product, then?
Your product, for $n$ even has value:
$$\frac{n!!}{(n-1)!!},$$
and
$$\frac{(n-1)!!}{(n)!!},$$
for $n$ odd.
where $!!$ is a double-factorial.
$\prod_n f(n)$ converges iff $\sum_n \ln f(n)$ converges. You have that:
$$\sum_{n=1}^N\ln f(n)=(-1)^{N\pmod{2}}\sum_{n=1}^{N/2}[\ln(2n-1)-\ln(2n)],$$
where the sign outside depends on whether $N$ is even or odd.
However for large $n$,
$$[\ln(2n-1)-\ln(2n)]\approx -\frac{1}{2n},$$
so the series, and therefore the product, does not converge so it doesn't make sense to talk about it's value at $\infty$.