Find the limit of $ x_n = \prod_{j=2}^{n} \left(1 - \frac{2}{j(j+1)}\right)^2$ I am stuck on the following problem:  

Let $x_n=(1-\frac{1}{3})^2(1-\frac{1}{6})^2(1-\frac{1}{10})^2 \ldots...(1-\frac{1}{n(n+1)/2})^2, \text{where} \space n \geq 2$. Then $\lim_{n \to \infty}x_n=?$

I see that $x_n^{\frac{1}{2}}=\frac{2}{3}\frac{5}{6}\frac{9}{10} \ldots..\frac{(n-1)(n+2)}{n(n+1)}$. now not sure what to do next ?Any idea?
 A: Let $f(n)=\frac{n-1}{n+1}$. Then show $$1-\frac{1}{n(n+1)/2} = \frac{f(n)}{f(n+1)}$$
A: $$
\begin{align}
\prod_{j=2}^n\left(1-\frac2{j(j+1)}\right)
&=\prod_{j=2}^n\left(\frac{(j-1)(j+2)}{j(j+1)}\right)\\
&=\color{#00A000}{\prod_{j=2}^n\frac{j-1}{j}}
\color{#C00000}{\prod_{j=2}^n\frac{j+2}{j+1}}\\
&=\color{#00A000}{\frac1n}\color{#C00000}{\frac{n+2}{3}}\\
&=\frac13\frac{n+2}{n}
\end{align}
$$
Taking it to the limit, we get
$$
\prod_{j=2}^\infty\left(1-\frac2{j(j+1)}\right)=\frac13
$$
The product of the squares is the square of the product (to get the answer to the modified question).
A: Note you really have a telescoping product. To see this intuitively, write out a few terms. You start with $\frac{1 \cdot 4}{2 \cdot 3}$
$$\frac{1 \cdot 4}{2 \cdot 3} \times \frac{2 \cdot 5}{3 \cdot 4} = \frac{1 \cdot 5}{3 \cdot 3}$$
and adding another term
$$\frac{1 \cdot 5}{3 \cdot 3} \times \frac{3 \cdot 6}{4 \cdot 5} = \frac{1 \cdot 6}{3 \cdot 4}$$
and proceed by induction to prove that $x_n = \frac{1 \cdot (n+2)}{3 \cdot n}$ which in the limit should converge to 1/3.
