I ran across this infinite product:


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


Then, here is my question


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

  • $\begingroup$ 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. $\endgroup$ – Ross Millikan Mar 26 '11 at 20:28

$\binom{k+1}{2}=\frac{(k+1)k}{2}$, 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.

  • $\begingroup$ Thank you. I knew I was overlooking something. I wasn't seeing the forest for the trees type situation. Thanks much. $\endgroup$ – 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.

  • $\begingroup$ Thank you. I see much better now. $\endgroup$ – Cody Mar 26 '11 at 20:47

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