# Thread: Proof (product of fractions)

1. ## Proof (product of fractions)

I need to prove that the product of 1/2 * 3/4 * 5/6 . . . 99/100 is less than 1/10. I understand that the product will continue to approach zero at a decreasing rate as the series progresses, but I need a mathematical explanation (of the most simple terms possible) as to why the product is less than 1/10.

Thank you very much, this problem has been driving me crazy. I'm in calculus 3, so this may be in the wrong section, but calculus 2 is the bound of my knowledge. If at all possible, lets keep it that simple. If not, blow me away anyways.

2. It's possible to prove this by proving a stronger result by induction - that $\displaystyle \displaystyle \frac{1 \cdot 3 \cdots \cdot (2n-1)}{2 \cdot 4 \cdots \cdot (2n)} \le \frac{5}{n}$
Can you try to do that?

3. I'm not quite sure how the 5/n came about in there to be completely honest.

4. Well, if you put n=50 you get the desired result. I just noticed that and tried to prove a more general statement using induction.

5. I'm sorry to sound incompetent but would you mind explaining a little more? I'm unfamiliar with the term "induction" and I am uncertain as to how the equation you gave proves the product of those fractions is less than 1/10.

6. Induction is a technique used to prove statements like this one - Mathematical induction - Wikipedia, the free encyclopedia. Are you sure you haven't heard of it?
Assuming you proved the inequality I gave for every $\displaystyle n \in \mathbb{N}$, if you put n = 50 you get the desired result.

7. Unfortunately I have not, at least not in that terminology. Nor have I dealt with the three symbols you mentioned (n E N).

8. If I set the term "1/2 * 3/4 * 5/6. . . 99/100" equal to A and create term B which equals "2/3 * 4/5 * 6/7. . . 98/99". This being the case, A < B. A * B = 1/100, and as 1/100 is 1/10 squared I have to prove that A^2 is less than 1/100. if A * B = 1/100 and A < B then A * A must be less than 1/100 and thus A < 1/10.

Does that work as a proof? I figured using a complimentary term might be easiest and as far as I can tell this works out. . .

9. That's nice.
You'll have to explain why A < B, however (for the proof, I mean)