# A finite product : $\prod_{k=0}^{n-1}(1-\frac{1}{n-1+k})$

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 ?

• Can you fix your typos please? Jul 7, 2019 at 12:59

HINT: (Assuming $$n \geq 2$$)

$$\prod_{k=0}^{n-1}(1-\frac{1}{n+k-1}) = \prod_{k=0}^{n-1}\frac{n+k-2}{n+k-1} = \frac{n-2}{n+(n-1)-1} = \frac{1}{2} \cdot \frac{n-2}{n-1}$$

• Nice solution sir @Presage , can you see the second product ? Jul 7, 2019 at 13:11
• @TomasHoubaze You do not need to use "sir" in addressing everyone! Jul 7, 2019 at 13:16

Hint: Prove by induction that $$\prod_{k=0}^{n-1}1-\frac{1}{n-1+k}=\frac{n-2}{2 (n-1)}$$

• Thanks @Dr. Sonnahard Graubner , can you see the product $B$ Jul 7, 2019 at 13:14
• @Dr.SonnhardGraubner: Your left-hand side is $1-\frac{1}{n-1+k}$ without brackets. Jul 11, 2019 at 22:17

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.

• Thank you sir @Adam Latosiñski , sir can telle this $\prod_{k=0}^{n-1}(2(n-k)+1)$=? Jul 7, 2019 at 14:05
• @TomasHoubaze This one is more complicated, brcause it contains $2k$, which means that it goes every second number. Such priducts are denoted using $!!$ symbol, that is $$1\cdot 3\cdot 5\cdot \dots \,\cdot (2n+1) = (2n+1)!!$$ $$2\cdot 4\cdot 6\cdot \dots \,\cdot (2n) = (2n)!!$$ They can be written using usual factorials: $$(2n)!! = 2^n n!$$ $$(2n+1)!! = \frac{(2n+1)!}{(2n)!!} = \frac{(2n+1)!}{2^n n!}$$ using the double factorial this product can be calculated. Jul 7, 2019 at 14:07
• Thank you very much sir @Adam Jul 7, 2019 at 14:10
• Can you please complete !!, so product equal : $(2n+1)!$ Jul 7, 2019 at 14:12
• $\prod_{k=0}^{n-1}(2n-2k+1) = (2n+1)!!$, if that's what you're asking about. Jul 7, 2019 at 23:16