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?

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.
