What is this product? I'm just curious because I was trying to come up with a weird function with weird discontinuities.  Then I thought
$$f(x)=\prod_{n=1}^\infty \dfrac{1}{1-\frac{1}{nx}}$$
So what's this product?
 A: 
This was an answer to the OP's first question which has since been edited, which asked about the product $\displaystyle \prod_{n=1}^{\infty} \dfrac{n}{nx-1}$.

Consider the possible values of $x$ in cases.


*

*If $|x|>1$, then for sufficiently large $n$, you'll have $|nx-1|>a|n|$ for some $a$ with $|a|>1$, so for these values of $x$, $f(x) = 0$.

*For $-1 \le x < 0$, we have
$$\dfrac{n}{nx-1} = -\dfrac{n}{n|x|+1} \to -\dfrac{1}{|x|} \le -1$$
and so the product doesn't converge.

*Writing $\dfrac{n}{nx-1} = \dfrac{1}{x-\frac{1}{n}}$ makes it clear that for any $0 \le x < 1$ the function diverges, since you're taking the product of an infinite sequence of numbers $>1$ which does not converge to $1$.

*Finally, $f(1)$ is undefined since one of the terms of your product is $\dfrac{1}{x-1}$.
In summary: $f(x)=0$ for $|x|>1$ and is undefined for $|x|<1$.
A: For amended question. When $x >1,$ the log of the partial product to $k$ terms
is $\sum_{n=1}^{k} \log(1 - \frac{1}{nx}),$ Let $r$ be the smallest integer with 
$rx >1.$ Then the partial product is greater than $\sum_{n=r}^{k} \frac{1}{nx}$ which tends to infinity as $k$ increases, since the harmonic series diverges.
Hence the product diverges for $x > 1.$
 If $x = \frac{1}{n}$ for any positive integer $n,$ then the product is not defined
for that $x.$
 For other values of $x <1,$ the product diverges by an argument similar to that at the beginning.
