What's the $\lim_{m\to\infty}\prod_{k=1}^m (1-e^{-kn})$?

Question

I try to show $\lim_{n\to\infty}\lim_{m\to\infty}\prod_{k=1}^m (1-e^{-kn})=1$. It seems we need to give a lower bound of $\lim_{m\to\infty}\prod_{k=1}^m (1-e^{-kn})$ depending on $n$ and as $n$ tends to infinity this lower bound tends to 1. I am trying to calculate $\log(\prod_{i=1}^m (1-e^{-in}))$ and see if it is closed to 0 with the fact that $\log(1-x)\approx -x$ as $x\to 0$. But I am not sure how to control the error.

Answer 1

As you have mentioned, it is easy to see that the limit is smaller than $1$ (as each factor in the product is smaller than 1).

Taking the logarithm is also a good idea. We have $$\log\left(\prod_{k=1}^m (1-e^{-k n})\right) =\sum_{k=1}^m \log(1-e^{-k n}) \geq \int_0^\infty \!dx\,\log(1-e^{-x n})= -\frac{\pi^2}{6n} . $$ as $\log(1-e^{-k n})< 0$ with monotonously decreasing absolute value; see here. It follows that $$ 1\geq \lim_{n\to\infty} \lim_{m\to\infty}\prod_{k=1}^m (1-e^{-k n}) \geq \lim_{n\to\infty} e^{-\pi^2/6 n} =1. $$

Answer 2

$$\displaystyle \lim_{m\to\infty}\prod_{k=1}^m (1-e^{-kn}) \geq \lim_{m\to\infty}1 - \sum_{k = 1}^me^{-kn} = 1 - \frac{e^{-n}}{e^{-n} - 1} = \frac{e^n - 2}{e^n - 1}$$