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

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Limits

lim_{n \to \infty} {(2^{\frac{1}{2}} - 2^{\frac{1}{3}}) (2^{\frac{1}{2}} - 2^{\frac{1}{5}}) \dots (2^{\frac{1}{2}} - 2^{\frac{1}{2 n + 1}})}

is equal to

Answer: (D)

Each factor

a_{k} = 2^{1/2} - 2^{1/ (2 k + 1)}

is positive, since

2^{1/2} > 2^{1/ (2 k + 1)}

for

k \geq 1

. Because

2^{1/ (2 k + 1)} > 1

, every factor obeys

a_{k} < 2^{1/2} - 1 < 1

. Hence the product of the

n

factors satisfies

0 < k = 1 \prod n a_{k} < (2^{1/2} - 1)^{n}

. As

2^{1/2} - 1 \approx 0.414

lies in

(0, 1)

(2^{1/2} - 1)^{n} \to 0

n \to \infty

, so by the squeeze theorem the value of the product is

0

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Single correct option

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