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Find the maximum and minimum of the following products :
$A)$ $\prod_{k=0}^{n-1}(1-\frac{1}{n-1+k})$
$B)$ $\prod_{k=0}^{n-1}(1-\frac{1}{n+1-k})$
My idea is :
$n-1+k>k$ then : $\frac{1}{n-1+k}<\frac{1}{k}$
We obtain :
$\prod_{k=0}^{n-1}(1-\frac{1}{k})$
But I don't have ideas to complete my work , and is my attempt correct ?
HINT: (Assuming $ n \geq 2 $)
$\prod_{k=0}^{n-1}(1-\frac{1}{n+k-1}) = \prod_{k=0}^{n-1}\frac{n+k-2}{n+k-1} = \frac{n-2}{n+(n-1)-1} = \frac{1}{2} \cdot \frac{n-2}{n-1}$
Hint: Prove by induction that $$\prod_{k=0}^{n-1}1-\frac{1}{n-1+k}=\frac{n-2}{2 (n-1)}$$
Let us note that for $a,b\in\mathbb N$, $2\le a \le b$ we have $$ \prod_{m=a}^b m = \frac{\prod_{m=1}^b m}{\prod_{m=1}^{a-1} m} = \frac{b!}{(a-1)!}$$
For the product A) we have $$ \prod_{k=0}^{n-1} \left(1-\frac{1}{n-1+k}\right) = \prod_{k=0}^{n-1}\frac{n-2+k}{n-1+k} = \frac{\prod_{k=0}^{n-1}(n-2+k)}{\prod_{k=0}^{n-1}(n-1+k)} = \frac{\prod_{m=n-2}^{2n-3} m}{\prod_{m=n-1}^{2n-2} m} = \\ = \frac{\frac{(2n-3)!}{(n-3)!}}{\frac{(2n-2)!}{(n-2)!}} = \frac{(n-2)! \cdot (2n-3)!}{(n-1)! \cdot (2n-2)!} = \frac{n-2}{2n-2}$$
For the product B) we have $$ \prod_{k=0}^{n-1} \left(1-\frac{1}{n+1-k}\right) = \prod_{k=0}^{n-1}\frac{n-k}{n+1-k} = \frac{\prod_{k=0}^{n-1}(n-k)}{\prod_{k=0}^{n-1}(n+1-k)} = \frac{\prod_{m=1}^{n} m}{\prod_{m=2}^{n+1} m} = \\ = \frac{m!}{\frac{(m+1)!}{1!}} = \frac{1! \cdot n!}{(n+1)!} = \frac{1}{n+1}$$
You can use such methods to solve other similar products.
