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Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
The value of the limit ∏n=2∞(1−1n2)\displaystyle\prod_{n=2}^{\infty}\left(1-\dfrac{1}{n^{2}}\right) is equal
A11
B14\dfrac{1}{4}
C13\dfrac{1}{3}
D12\dfrac{1}{2}correct
Solution
Step 1: Factor: 1−1n2=n2−1n2=(n−1)(n+1)n⋅n1-\dfrac{1}{n^{2}}=\dfrac{n^{2}-1}{n^{2}}=\dfrac{(n-1)(n+1)}{n\cdot n}. Step 2: Write the partial product up to NN:
∏n=2Nn−1n⋅n+1n=(∏n=2Nn−1n)(∏n=2Nn+1n)\prod_{n=2}^{N}\dfrac{n-1}{n}\cdot\dfrac{n+1}{n}=\left(\prod_{n=2}^{N}\dfrac{n-1}{n}\right)\left(\prod_{n=2}^{N}\dfrac{n+1}{n}\right)
 Step 3: Each is telescoping:
∏n=2Nn−1n=12⋅23⋅34⋯N−1N=1N\prod_{n=2}^{N}\dfrac{n-1}{n}=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdots\dfrac{N-1}{N}=\dfrac{1}{N}
∏n=2Nn+1n=32⋅43⋅54⋯N+1N=N+12\prod_{n=2}^{N}\dfrac{n+1}{n}=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdots\dfrac{N+1}{N}=\dfrac{N+1}{2}
 Step 4: Multiply:
∏n=2N(1−1n2)=1N⋅N+12=N+12N\prod_{n=2}^{N}\left(1-\dfrac{1}{n^{2}}\right)=\dfrac{1}{N}\cdot\dfrac{N+1}{2}=\dfrac{N+1}{2N}
 Step 5: Take N→∞N\to\infty:
lim⁡N→∞N+12N=12\lim_{N\to\infty}\dfrac{N+1}{2N}=\dfrac{1}{2}
 Correct answer: (4) 12\dfrac{1}{2}
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