The value of the infinite product prodlimits_ | vedclass

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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

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$\frac{1}{2}$

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

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