Is there a known upper bound on $\prod\limits_{i=1}^{n}(1-ix)$

Question

It seems intuitive that $(1-\epsilon)^n \geq (1-\epsilon) \cdot (1-2\epsilon) \cdot \ldots \cdot (1- n \epsilon)$, but not by how much or how significantly the difference between these grows dependent on $n$, if at all.

It's known that $(1-\epsilon)^n \leq \frac{1}{1+n\epsilon}$ for $0 < \epsilon < 1$ and $n > 1$. Is there a similar, but tighter, upper bound on the series $\displaystyle \prod_{i=1}^{n} \left(1-i\epsilon\right)$ under these conditions? Or even better, are there direct lower or upper bounds on the difference $\left(\displaystyle \prod_{i=1}^{n} \left( 1-i\epsilon \right)\right) - (1-\epsilon)^n$?

Answer 1

The product vanishes as $n\to\infty$, for $\varepsilon\in(0,\frac{1}n)$, which is covered in Marty Cohen's answer. It also seems to explode for $\varepsilon\in(0.5,1)$, which I will describe. But some strange oscillations seem to occur for $\varepsilon\in(\frac1n,0.5)$.

Let $\varepsilon\in(0.5,1)$. Then $\prod_{i=1}^{n}\left(1-i\varepsilon\right)=\left(1-\varepsilon\right)\cdot\exp\left[\sum_{i=2}^{n}\ln\left(i\varepsilon-1\right)\right]$. We don't include the $i=1$ term in the sum because this is the only positive term over $(0.5,1)$; we invert the sign of the other terms in order to take their logarithm.

Then, given $\ln x\leq x-1$ and $\sum_{i=1}^ni=\frac{n(n+1)}2$, the modulus of the product is bounded by

$$\left(1-\varepsilon\right)\exp\left[\frac{\left(n-1\right)\left(n+2\right)}{2}\varepsilon-2\left(n-1\right)\right]$$

The sign of the product here is simply $(-1)^{n+1}$. The same technique can create a bound for $(1,\infty)$.

Your conjectured bound of $(1-\varepsilon)^n\geq\prod_{i=1}^{n}\left(1-i\varepsilon\right)$ is incorrect, as shown by $n=3, \varepsilon=0.8$. However it is true for $n\geq4, \varepsilon\in(0,\frac1n)$ as it transitively bounds Marty Cohen's bound.

Answer 2

Let $X = \displaystyle \prod_{i=1}^{n}(1-ix)$ and let $Y = \log(X)$. More specifically:

$$ Y = \log\left(\prod_{i=1}^{n}(1-ix)\right) = \sum_{i=1}^n \log(1-ix).$$

If $x > 0$, it is clear that $$\max_x (1-ix) = 1.$$

Therefore:

$$(1-ix) \leq 1 \Rightarrow \log(1-ix) \leq 0 \Rightarrow Y \leq 0 \Rightarrow e^Y \leq e^0 \Rightarrow X \leq 1.$$

Concluding:

$$\displaystyle \prod_{i=1}^{n}(1-ix) \leq 1,$$

for $x > 0$ implies that the upper-bound you are looking for is $1$.

Remark

This calculations work since both $e^{(...)}$ and $\log(...)$ are monotonically increasing functions.

Answer 3

Here is some naive plugging in which depends on $|nx| < 1$. I don't know how useful it is.

$\begin{array}\\ f(n, x) &= \prod_{k=1}^{n} (1-kx)\\ g(n, x) &=\ln(f(n, x))\\ &= \sum_{k=1}^{n} \ln(1-kx)\\ &= -\sum_{k=1}^{n} \sum_{j=1}^{\infty} \dfrac{(kx)^j}{j}\\ &= -\sum_{j=1}^{\infty}\dfrac1{j} \sum_{k=1}^{n} (kx)^j\\ &= -\sum_{j=1}^{\infty}\dfrac{x^j}{j} \sum_{k=1}^{n} k^j\\ &\approx -\sum_{j=1}^{\infty}\dfrac{x^j}{j} \left(\dfrac{n^{j+1}}{j+1}+\dfrac{n^j}{2}\right) \qquad\text{first two terms of the expansion}\\ &= -\sum_{j=1}^{\infty}\dfrac{x^j}{j}\dfrac{n^{j+1}}{j+1}-\sum_{j=1}^{\infty}\dfrac{x^j}{j}\dfrac{n^j}{2}\\ &= -n\sum_{j=1}^{\infty}\dfrac{(nx)^j}{j(j+1)}-\dfrac12\sum_{j=1}^{\infty}\dfrac{(nx)^j}{j}\\ &= -n\left(1+\dfrac{(1-nx)\ln(1-nx)}{nx}\right)+\dfrac{\ln(1-nx)}{2} \qquad\text{if } |nx| < 1\\ &= -n-\dfrac{(1-nx)\ln(1-nx)}{x}+\dfrac{\ln(1-nx)}{2}\\ &= -n-\ln(1-nx)\left(\dfrac{(1-nx)}{x}+\dfrac12\right)\\ &= -n-\ln(1-nx)\left(\dfrac1{x}-n+\dfrac12\right)\\ \end{array} $