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 :

Solution

Maths sir · 4 steps

Do not multiply anything out: trap each bracket between 0 and a fixed number below 1, then squeeze.

1. Size of one factor. The general factor is 2^(1/2) - 2^(1/(2k+1)) for k = 1, 2, …, n. Since 0 < 1/(2k+1) < 1/2 for every k ≥ 1, we get 1 < 2^(1/(2k+1)) < √2.
2. Bound it. So every factor is positive and less than √2 - 1 ≈ 0.414. Multiplying all n factors gives the bound below for the product Pₙ.

   0 < Pₙ < (√2-1)ⁿ

3. Squeeze. Because 0 < √2 - 1 < 1, the power (√2-1)ⁿ → 0 as n → ∞. Pₙ sits between 0 and something tending to 0, so Pₙ → 0 as well.
4. So. The limit is 0, which is option (d).

The idea to keep. For a limit of a growing product, check the size of each factor. If every factor stays between 0 and a fixed number less than 1, the product goes to 0 by the sandwich theorem.

Answer from the NTA final answer key. Solution by PrepWiser.