\(lim _{n→∞}{(2^½-2^½)(2^½-2^½)….(2^½-^2{\frac{1}{2n+1}})}\) is equal to
- 1
- 0
- \(\frac{1}{\sqrt2}\)
- \(\sqrt2\)
The Correct Option is B
Solution and Explanation
First, we analyze the behavior of each term in the product. The smallest term in the product is: \[ \frac{1}{2^2 - 2^3} \quad \text{and the largest term is} \quad \frac{1}{2^2 - 2^{2n+1}} \] The product is bounded as: \[ \left( \frac{1}{2^2 - 2^3} \right)^n \leq P \leq \left( \frac{1}{2^2 - 2^{2n+1}} \right)^n \] The sequence is bounded between 0 and 1. Therefore, the limit of the product as \( n \to \infty \) is 0. Thus, the final answer is: \[ P = 0 \]
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