Evaluating $\prod\limits_{n\ge2}\left(1-\frac2{n(n+1)}\right)^2$

Question

Let $$x_n=\left(1-\frac13\right)^2\left(1-\frac16\right)^2\left(1-\frac1{10}\right)^2\cdots\left(1-\frac2{n(n+1)}\right)^2,n\ge2$$

Then the value of $\lim_{n\to\infty}x_n$ equals?

I tried to take log both sides: $$\log x_n=2\sum\limits_{k=2}^{n}\log⁡(1-\frac{2}{k(k+1)}) $$ Despite taking limit both sides, I did not get it of the standard form of $1^{\infty}$.

Answer 1

Hint:

Rewriting $$1-\frac2{k(k+1)}=\frac{(k-1)(k+2)}{k(k+1)}=\frac{k-1}{k}\cdot\frac{k+2}{k+1}$$

Thus $$\prod_{k=2}^n\left[1-\frac2{k(k+1)}\right]=\prod_{k=2}^n\frac{k-1}{k}\cdot\prod_{k=2}^n\frac{k+2}{k+1}$$