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