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

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 $\color{red}{\large\frac{1}{9}}$.
