lim n { ( 2 1 2 - 2 1 3 ) ( 2 1 2 - 2 1 5 ) ? ( 2 1 2 - 2 1 2 n + 1 ) } is equal to [2023]

Question

$\lim_{n \to \infty} {(2^{\frac{1}{^{2}}} - 2^{\frac{1}{^{3}}}) (2^{\frac{1}{^{2}}} - 2^{\frac{1}{^{5}}}) \dots (2^{\frac{1}{^{2}}} - 2^{\frac{1}{^{2 n + 1}}})}$ is equal to [2023]

Smart Achievers · Accepted Answer

(3)

Let $L = \lim_{n \to \infty} {(2^{1 / 2} - 2^{1 / 3}) (2^{1 / 2} - 2^{1 / 5}) \dots (2^{1 / 2} - 2^{1 / (2 n + 1)})}$

By Sandwich Theorem,

${(2^{1 / 2} - 2^{1 / 3})}^{n} < (2^{1 / 2} - 2^{1 / 3}) (2^{1 / 2} - 2^{1 / 5}) (2^{1 / 2} - 2^{1 / 7}) \dots (2^{1 / 2} - 2^{1 / (2 n + 1)}) < {(2^{1 / 2} - 2^{1 / (2 n + 1)})}^{n}$

$\Rightarrow \lim_{n \to \infty} {(2^{1 / 2} - 2^{1 / 3})}^{n} < L < \lim_{n \to \infty} {(2^{1 / 2} - 2^{1 / (2 n + 1)})}^{n}$

As, $\lim_{n \to \infty} {(2^{1 / 2} - 2^{1 / 3})}^{n} = 0 and \lim_{n \to \infty} {(2^{1 / 2} - 2^{1 / (2 n + 1)})}^{n} = 0$

$∴$ $L = 0$

Solution

Q.
$\lim_{n \to \infty} {(2^{\frac{1}{^{2}}} - 2^{\frac{1}{^{3}}}) (2^{\frac{1}{^{2}}} - 2^{\frac{1}{^{5}}}) \dots (2^{\frac{1}{^{2}}} - 2^{\frac{1}{^{2 n + 1}}})}$ is equal to [2023]

1 $\frac{1}{\sqrt{2}}$

2 1

3 0

4 $\sqrt{2}$

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Solution

Q. limn→∞{(212-213)(212-215)⋯(212-212n+1)} is equal to [2023]

1 12

2 1

3 0

4 2

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Q.
$\lim_{n \to \infty} {(2^{\frac{1}{^{2}}} - 2^{\frac{1}{^{3}}}) (2^{\frac{1}{^{2}}} - 2^{\frac{1}{^{5}}}) \dots (2^{\frac{1}{^{2}}} - 2^{\frac{1}{^{2 n + 1}}})}$ is equal to [2023]

1 $\frac{1}{\sqrt{2}}$

4 $\sqrt{2}$