The value of the infinite product prodlimits_ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The given expression is $\prod\limits_{n = 2}^\infty  {\left( {\frac{{{n^2} - 1}}{{{n^2}}}} \right)} = \prod\limits_{n = 2}^\infty  {\left( {\frac

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(D) The given expression is $\prod\limits_{n = 2}^\infty  {\left( {\frac{{{n^2} - 1}}{{{n^2}}}} ight)} = \prod\limits_{n = 2}^\infty  {\left( {\frac{{(n - 1)(n + 1)}}{{n \cdot n}}} ight)}$.Expanding the product for $n=2$ to $N$:$P_N = \left( \frac{1 \cdot 3}{2 \cdot 2} ight) \cdot \left( \frac{2 \cdot 4}{3 \cdot 3} ight) \cdot \left( \frac{3 \cdot 5}{4 \cdot 4} ight) \cdots \left( \frac{(N-1)(N+1)}{N \cdot N} ight)$.This is a telescoping product:$P_N = \left( \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac{N-1}{N} ight) \cdot \left( \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdots \frac{N+1}{N} ight)$.$P_N = \left( \frac{1}{N} ight) \cdot \left( \frac{N+1}{2} ight) = \frac{N+1}{2N}$.Taking the limit as $N 	o \infty$:$\lim_{N 	o \infty} \frac{N+1}{2N} = \lim_{N 	o \infty} \frac{1 + 1/N}{2} = \frac{1}{2}$.

The value of the infinite product $\prod\limits_{n = 2}^\infty {\left( {1 - \frac{1}{{{n^2}}}} \right)}$ is

Similar Questions

$\sum_{k=0}^{12} \frac{1}{\sin \left((k+1) \frac{\pi}{6}+\frac{\pi}{4}\right) \sin \left(\frac{k \pi}{6}+\frac{\pi}{4}\right)} = $

The value of $\sum\limits_{k = 1}^\infty {\frac{{3{k^2} + 3k + 1}}{{{{\left( {{k^2} + k} \right)}^3}}}} $ is equal to

The sum of the series $\frac{1}{x+1}+\frac{2}{x^{2}+1}+\frac{2^{2}}{x^{4}+1}+\ldots+\frac{2^{100}}{x^{2^{100}}+1}$ when $x=2$ is:

The sum of the series $\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{n^2 - 1} + \sqrt{n^2}}$ equals

The sum $\sum_{n=1}^{21} \frac{3}{(4n-1)(4n+3)}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module