A confusing sequence of products

Question

Define

$x_n=\left(1-\frac{1}{3}\right)^2\left(1-\frac{1}{6}\right)^2\left(1-\frac{1}{10}\right)^2\left(1-\frac{1}{15}\right)^2...\left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^2$ for $n>1$

then $\lim_{n\to\infty} x_n $ is

A)..$\frac{1}{3}$

B)..$\frac{1}{9}$

C)..$\frac{1}{81}$

D)..$0$

What I did is...

$X_n$=$(\prod_{i=2}^{i} \frac{i^2+i-2}{i(i+1)})^2$=$(\prod_{i=2}^{i} \frac{(i+2)(i-1)}{i(i+1)})^2$=$(\frac{4.5.6.....(i+2)()1.2.3.4....(i-1)}{(1.2.3.4...i)(2.3.4.5...(i+1))})^2=(\frac{1}{3}(1+\frac{2}{i}))^2$

as $i\to\infty$ we get $\lim_{x\to\infty} x_n=\frac{1}{9}$

Am I correct or wrong? Any help would be appreciated.

Answer 1

\begin{eqnarray*} \frac{ 1 \times \color{red}{2}} {3 \times \color{orange}{1}} \frac{ \color{orange}{1} \times \color{blue}{5} } { \color{red}{2} \times \color{green}{3}} \frac{ \color{green}{3} \times \color{red}{3}} { \color{blue}{5} \times \color{orange}{2}} \frac{ \color{orange}{2}\times \color{blue}{7} } { \color{red}{3} \times \color{green}{5}} \cdots \end{eqnarray*} Square what is left & you are right the answwer is $\color{red}{1/9}$.