Finding the limit $\lim_{n\rightarrow\infty}(1-\frac{1}{3})^2(1-\frac{1}{6})^2(1-\frac{1}{10})^2...(1-\frac{1}{n(n+1)/2})^2$

Question

Any hints on how to find the following limit ?.

I haven't been able to figure it out still. $$ \lim_{n \to \infty}\left(\, 1 - \frac{1}{3}\, \right)^{2} \left(\, 1 - \frac{1}{6}\, \right)^{2}\left(\, 1 - \frac{1}{10}\, \right)^{2}\ldots\left[\, 1 - \frac{1}{n\left(\, n + 1\, \right)/2}\, \right]^{2} $$

Pardon me if it's a silly question.

Edit: Solved upto this much

$$ \prod_{n = 2}^{\infty}\left[\, 1-\frac{2}{n\left(\, n + 1\, \right)}\, \right] = \prod_{n = 2}^{\infty}\left(\, \frac{n + 2}{n + 1}\,\frac{n - 1}{n}\, \right) $$

I did'nt notice most of the terms cancel out.

Answer 1

We have $$ \prod_{n=2}^{+\infty}\left(1-\frac{2}{n(n+1)}\right)=\prod_{n=2}^{+\infty}\left(\frac{n+2}{n+1}\cdot\frac{n-1}{n}\right)=\frac{1}{3} $$ (it is a telescopic product) hence you limit equals $\color{red}{\large\frac{1}{9}}$.

Answer 2

It may be interesting to see that, using the the Gamma function in the form $$\Gamma\left(z\right)=\lim_{n\rightarrow\infty}\frac{n^{z-1}n!}{z\left(z+1\right)\cdots\left(z+n-1\right)}$$ we can prove that $$\prod_{n\geq0}\frac{\left(n+a\right)\left(n+b\right)}{\left(n+c\right)\left(n+d\right)}=\frac{\Gamma\left(c\right)\Gamma\left(d\right)}{\Gamma\left(a\right)\Gamma\left(b\right)},\, a+b=c+d $$ so $$\prod_{n\geq2}\frac{\left(n+2\right)\left(n-1\right)}{\left(n+1\right)n}=\prod_{n\geq0}\frac{\left(n+4\right)\left(n+1\right)}{\left(n+3\right)\left(n+2\right)}$$ $$=\frac{\Gamma\left(3\right)\Gamma\left(2\right)}{\Gamma\left(4\right)\Gamma\left(1\right)}=\color{red}{\frac{1}{3}}.$$