Question

Let $x_{n}=\left(1-\frac{1}{3}\right)^{2}\left(1-\frac{1}{6}\right)^{2}\left(1-\frac{1}{10}\right)^{2} ........ \left(1-\frac{1}{\frac{n\left(n+1\right)}{2}}\right)^2, n \ge 2.$ Then the value of $\displaystyle \lim_{n \to \infty} x_n$ is

• A. 44564
• B. 44570
• C. Jan-81
• D. 0

Answer 1

The Correct Option is B

We have, $x_{n}=\left[\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\right.$
$\left.\left(1-\frac{1}{10}\right) \ldots\left(1-\frac{2}{n(n+1)}\right)\right]^{2}$
$\Rightarrow x_{n} =\left[\prod_{n=2}^{n}\left(\frac{n^{2}+n-2}{n(n+1)}\right)\right]^{2}$
$=\left[\prod_{n=2}^{n}\left(\frac{(n+2)(n-1)}{n(n+1)}\right)\right]^{2}$
$=\left[\prod_{n=2}^{n}\left(\frac{n+2}{n+1}\right) \cdot \prod_{n=2}^{n}\left(\frac{n-1}{n}\right)\right]^{2}$
$=\left[\prod_{n=2}^{n}\left(\frac{n+2}{n+1}\right)\right]^{2}\left[\prod_{n=2}^{n}\left(\frac{n-1}{n}\right)\right]^{2}$
$\Rightarrow x_{n}=\left(\frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \ldots \frac{n+2}{n+1}\right)^{2}\left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \ldots \frac{n-1}{n}\right)^{2}$
$\Rightarrow x_{n}=\left(\frac{n+2}{3}\right)^{2}\left(\frac{1}{n}\right)^{2}$
$\Rightarrow x_{n}=\frac{1}{9}\left(\frac{n+2}{n}\right)^{2}$
$\Rightarrow x_{n}=\frac{1}{9}\left(1+\frac{2}{n}\right)^{2}$
$\Rightarrow \displaystyle\lim _{n \rightarrow \infty} x_{n}=\frac{1}{9}(1+0)^{2}=\frac{1}{9}$