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Mathematics Question on limits and derivatives

Let xn=(1−13)2(1−16)2(1−110)2........(1−1n(n+1)2)2,n≥2.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 lim⁡n→∞xn\displaystyle \lim_{n \to \infty} x_n is

A

44564

B

44570

C

Jan-81

D

0

Answer

44570

Explanation

Solution

We have, xn=[(1−13)(1−16)x_{n}=\left[\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\right.
(1−110)…(1−2n(n+1))]2\left.\left(1-\frac{1}{10}\right) \ldots\left(1-\frac{2}{n(n+1)}\right)\right]^{2}
⇒xn=[∏n=2n(n2+n−2n(n+1))]2\Rightarrow x_{n} =\left[\prod_{n=2}^{n}\left(\frac{n^{2}+n-2}{n(n+1)}\right)\right]^{2}
=[∏n=2n((n+2)(n−1)n(n+1))]2=\left[\prod_{n=2}^{n}\left(\frac{(n+2)(n-1)}{n(n+1)}\right)\right]^{2}
=[∏n=2n(n+2n+1)⋅∏n=2n(n−1n)]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}
=[∏n=2n(n+2n+1)]2[∏n=2n(n−1n)]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}
⇒xn=(43⋅54⋅65…n+2n+1)2(12⋅23⋅34…n−1n)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}
⇒xn=(n+23)2(1n)2\Rightarrow x_{n}=\left(\frac{n+2}{3}\right)^{2}\left(\frac{1}{n}\right)^{2}
⇒xn=19(n+2n)2\Rightarrow x_{n}=\frac{1}{9}\left(\frac{n+2}{n}\right)^{2}
⇒xn=19(1+2n)2\Rightarrow x_{n}=\frac{1}{9}\left(1+\frac{2}{n}\right)^{2}
⇒lim⁡n→∞xn=19(1+0)2=19\Rightarrow \displaystyle\lim _{n \rightarrow \infty} x_{n}=\frac{1}{9}(1+0)^{2}=\frac{1}{9}