Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I ran across this infinite product:

$$\lim_{n\to\infty}\prod_{k=2}^n\left(1-\frac1{\binom{k+1}{2}}\right)$$

I easily found that it converges to 1/3. Using my calculator, I found that

$$1-\frac1{\binom{k+1}{2}}=\frac{(k-1)(k+2)}{k(k+1)}$$

Then, here is my question

$$\prod_{k=2}^n\frac{(k-1)(k+2)}{k(k+1)}=\frac{n+2}{3n}$$

This is what my calculator gave me. How did it arrive at this? That is, how could I do this by hand if I wanted to? I tried writing out some terms and even the (n+1)st term, made cancellations, but it did not work out. I feel rather obtuse. How does one find a closed form for a partial infinite product like this?

$$\frac23\cdot \frac56\cdot \frac9{10}\cdot\cdot\cdot \frac{(n-1)(n+2)}{n(n+1)}\cdot \frac{n(n+3)}{(n+1)(n+2)}$$

Making the cancellations leaves $\frac{(n-1)(n+3)}{(n+1)^2}$, not $\frac{n+2}{3n}=\frac2{3n}+\frac13$.

This is why it converges to 1/3. It is easy to see the limit. That is not my concern.

It is how does one arrive at the closed form of $\frac{n+2}{3n}$ for this 'finite' product?

I am overlooking something obvious. I just know it.

Thank you

share|improve this question
    
You went one term too far in expanding the product-there is no $\frac{n(n+3)}{(n+1)(n+2)}$ Then there is an $(n+1)$ in the numerators from the last omitted term and both $(n-1)$'s cancel. Just a question of keeping track of all the terms. –  Ross Millikan Mar 26 '11 at 20:28
    
Thanks Ross. I see now. –  Cody Mar 26 '11 at 20:47

2 Answers 2

up vote 5 down vote accepted

$\binom{k+1}{2}=\frac{(k+1)k}{2}$, so $1-\frac{1}{\binom{k+1}{2}}=\frac{k(k+1)-2}{k(k+1)}=\frac{(k-1)(k+2)}{k(k+1)}$. Then $$\prod_{k=2}^{n}\frac{(k-1)(k+2)}{k(k+1)}=\frac{2(n-1)!(n+2)!}{6n!(n+1)!}=\frac{n+2}{3n}$$ where the 2 comes because the $(n+1)!$ in the denominator really starts at 3 and the 6 comes because the $(n+2)!$ really starts at 4. Then for the last equality we just recognize which terms don't cancel in the factorials.

Often when working on problems like this, it is not a good idea to reduce fractions. If your partial result had shown $\frac{1 \cdot 4}{2 \cdot 3}$ instead of $\frac{2}{3}$ and so on, it would have been easier to see the pattern.

share|improve this answer
    
Thank you. I knew I was overlooking something. I wasn't seeing the forest for the trees type situation. Thanks much. –  Cody Mar 26 '11 at 20:46

A much simpler example would be $\prod_{k=2}^n \frac{k}{k+1} = \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdots \frac{n}{n+1} = \frac{2}{n+1}$. All the other terms here, obviously, cancel. In your case, much cancels but it's not quite as obvious.

share|improve this answer
    
Thank you. I see much better now. –  Cody Mar 26 '11 at 20:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.