TITLE: How to find closed form for a partial infinite product?
QUESTION [5 upvotes]: 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

REPLY [5 votes]: $\binom{k+1}{2}=\frac{(k+1)k}{2}$, 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.