Question

Let $x_n={(1-\frac{1}{3})}^2{(1-\frac{1}{6})}^2{(1-\frac{1}{10})}^2...{(1-\frac{1}{\frac{n(n+1)}{2}})}^2,n\ge 2$. Then, the calculate of $\underset{n\to \infty }{lim}{X}_n$ is

• A. $\frac{1}{3}$
• B. $\frac{1}{9}$
• C. $\frac{1}{81}$
• D. $0$

Answer 1

Answer: (B)

$\frac{1}{9}$

We have,

$x_n={[(1-\frac{1}{3})(1-\frac{1}{6})(1-\frac{1}{10})...(1-\frac{2}{n(n+1)})]}^2$

$\Rightarrow x_n={\left[\prod _{n=2}^n\left(\frac{n^2+n-2}{n(n+1)}\right)\right]}^2$

$={\left[\prod _{n=2}^n\left(\frac{(n+2)(n-1)}{n(n+1)}\right)\right]}^2$

$={[\prod _{n=2}^n\left(\frac{n+2}{n+1}\right)\cdot \prod _{n=2}^n\left(\frac{n-1}{n}\right)]}^2$

$={\left[\prod _{n=2}^n\left(\frac{n+2}{n+1}\right)\right]}^2{\left[\prod _{n=2}^n\left(\frac{n-1}{n}\right)\right]}^2$

$\Rightarrow x_n={(\frac{4}{3}\cdot \frac{5}{4}\cdot \frac{6}{5}...)}^2{(\frac{1}{2}\cdot \frac{2}{3}\cdot \frac{3}{4}...\frac{n-1}{n})}^2$

$\Rightarrow x_n={\left(\frac{n+2}{3}\right)}^2{\left(\frac{1}{n}\right)}^2$

$\Rightarrow x_n=\frac{1}{9}{\left(\frac{n+2}{n}\right)}^2$

$\Rightarrow x_n=\frac{1}{9}{(1+\frac{2}{n})}^2$

$\Rightarrow \underset{n\to \infty }{lim}{x}_n=\frac{1}{9}(1+0{)}^2=\frac{1}{9}$