Under what conditions would the function $\prod_{i=1}^{n}{\frac{r_i}{r_i - 1}}$ be decreasing with respect to $n$?

Question

So I know that

$$\frac{r}{r - 1}$$

is a decreasing function of $r$.

My question is: Under what conditions would the following function be decreasing with respect to $n$?

$$\displaystyle \prod_{i=1}^{n}{\frac{r_i}{r_i - 1}}$$

Any hints would be much appreciated!

Answer 1

Suppose first that $\frac{r_1}{r_1-1}\lt 0$, that is, $r_1$ is between $0$ and $1$. Then for our product to be decreasing, we need $\frac{r_i}{r_i-1}\gt 1$ for all $i\ge 2$. This happens precisely if $r_i\gt 1$ for all $i\ge 2$.

Suppose next that $\frac{r_1}{r_1-1}\gt 0$. This can happen if $r_1\gt 1$ or $r_1\lt 0$. Then our product can be decreasing in two ways. It can stay positive for a while, or forever. It stays positive forever (but decreases) if for $i\ge 2$ we have $r_i\lt 0$.

Or else we can have $r_i\lt 0$ for $i=2$ to $n$, and $\frac{r_{n+1}}{r_{n+1}-1}$ negative (that is, $r_{n+1}$ between $0$ and $1$.) That makes the product negative, so for $j\gt n+1$ we must have $r_j\gt 1$.

Answer 2

$f'(r) < 0 \iff\dfrac{d}{dr}\ln(f(r)) < 0 \iff \dfrac{d}{dr}\displaystyle \sum_{i=1}^n\ln\left(\dfrac{r_i}{r_i-1}\right)=-\displaystyle \sum_{i=1}^n \dfrac{1}{r_i(r_i-1)}<0 \iff r_i > 1, \forall i \geq 1.$