jee-main 2023 Q72

jee-main · India · session2_06apr_shift2 Sequences and series, recurrence and convergence Convergence proof and limit determination

$\lim_{n \rightarrow \infty} \left\{\left(2^{\frac{1}{2}} - 2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}} - 2^{\frac{1}{5}}\right) \ldots \left(2^{\frac{1}{2}} - 2^{\frac{1}{2n+1}}\right)\right\}$ is equal to
(1) 1
(2) 0
(3) $\sqrt{2}$
(4) $\frac{1}{\sqrt{2}}$

$\lim_{n \rightarrow \infty} \left\{\left(2^{\frac{1}{2}} - 2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}} - 2^{\frac{1}{5}}\right) \ldots \left(2^{\frac{1}{2}} - 2^{\frac{1}{2n+1}}\right)\right\}$ is equal to\\
(1) 1\\
(2) 0\\
(3) $\sqrt{2}$\\
(4) $\frac{1}{\sqrt{2}}$

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