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Let $x_{n}=\left(1-\frac{1}{3}\right)^{2}\left(1-\frac{1}{6}\right)^{2}\left(1-\frac{1}{10}\right)^{2} \ldots$ $\left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^{2}, n \geq 2$ Then, the value of $\lim _{n \rightarrow \infty} x_{n}$ is
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$1 / 9$
We have, $x_{n}=\left[\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\right.$ $$
\left.\left(1-\frac{1}{10}\right) \ldots\left(1-\frac{2}{n(n+1)}\right)\right]^{2}
$$
$\begin{aligned} \Rightarrow x_{n} &=\left[\prod_{n=2}^{n}\left(\frac{n^{2}+n-2}{n(n+1)}\right)\right]^{2} \\ &=\left[\prod_{n=2}^{n}\left(\frac{(n+2)(n-1)}{n(n+1)}\right)\right]^{2} \\ &=\left[\prod_{n=2}^{n}\left(\frac{n+2}{n+1}\right) \cdot \prod_{n=2}^{n}\left(\frac{n-1}{n}\right)\right]^{2} \\ \Rightarrow &\left[\prod_{n=2}^{n}\left(\frac{n+2}{n+1}\right)\right]^{2}\left[\prod_{n=2}^{n}\left(\frac{n-1}{n}\right)\right]^{2} \\ \Rightarrow \quad x_{n}=&\left(\frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \ldots \frac{n+2}{n+1}\right)^{2}\left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \ldots \frac{n-1}{n}\right)^{2} \\ \Rightarrow \quad & x_{n}=\left(\frac{n+2}{3}\right)^{2}\left(\frac{1}{n}\right)^{2} \Rightarrow x_{n}=\frac{1}{9}\left(\frac{n+2}{n}\right)^{2} \\=& \quad \lim _{n \rightarrow \infty} x_{n}=\frac{1}{9}(1+0)^{2}=\frac{1}{9} \end{aligned}$
\left.\left(1-\frac{1}{10}\right) \ldots\left(1-\frac{2}{n(n+1)}\right)\right]^{2}
$$
$\begin{aligned} \Rightarrow x_{n} &=\left[\prod_{n=2}^{n}\left(\frac{n^{2}+n-2}{n(n+1)}\right)\right]^{2} \\ &=\left[\prod_{n=2}^{n}\left(\frac{(n+2)(n-1)}{n(n+1)}\right)\right]^{2} \\ &=\left[\prod_{n=2}^{n}\left(\frac{n+2}{n+1}\right) \cdot \prod_{n=2}^{n}\left(\frac{n-1}{n}\right)\right]^{2} \\ \Rightarrow &\left[\prod_{n=2}^{n}\left(\frac{n+2}{n+1}\right)\right]^{2}\left[\prod_{n=2}^{n}\left(\frac{n-1}{n}\right)\right]^{2} \\ \Rightarrow \quad x_{n}=&\left(\frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \ldots \frac{n+2}{n+1}\right)^{2}\left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \ldots \frac{n-1}{n}\right)^{2} \\ \Rightarrow \quad & x_{n}=\left(\frac{n+2}{3}\right)^{2}\left(\frac{1}{n}\right)^{2} \Rightarrow x_{n}=\frac{1}{9}\left(\frac{n+2}{n}\right)^{2} \\=& \quad \lim _{n \rightarrow \infty} x_{n}=\frac{1}{9}(1+0)^{2}=\frac{1}{9} \end{aligned}$
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