I want to show the following product converges for $x<e$ and diverges for $x \ge e$: $$\lim_{n\to\infty}\prod_{i=0}^{n-1}\left(x-\frac{xi}{n}\right).$$
To do this, I would need to show that the natural log of the product converges or diverges, i.e.: $$\lim_{n\to\infty}\sum_{i=0}^{n-1}\ln{\left(x-\frac{xi}{n}\right)}$$ converges or diverges. How would I show this?
The second expression equals $$ \lim_{n\to\infty} \bigg( n \bigg( \ln x + \sum_{i=0}^{n-1} \frac1n \ln\Big( 1-\frac in \Big) \bigg) \bigg). $$ The sum is a Riemann sum for $\int_0^1 \ln(1-x)\,dx =-1$. This isn't the end of the proof, because it's being multiplied by $n$ inside the limit, but it should suggest a way to finish off the proof.