$\lim _{n \rightarrow \infty} \frac{1}{2^{n}}\left(\frac{1}{\sqrt{1-\frac{1}{2^{n}}}}+\frac{1}{\sqrt{1-\frac{2}{2^{n}}}}+\frac{1}{\sqrt{1-\frac{3}{2^{n}}}}+\ldots+\frac{1}{\sqrt{1-\frac{2^{n}-1}{2^{n}}}}\right)$ is equal to

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Definite Integrals

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

is equal to

Answer: (C)

This is a Riemann sum. Let

x_{r} = \frac{r}{2 ^{n}}

where

r = 1, 2, \dots, 2^{n} - 1

. The sum becomes

\sum_{r = 1}^{2^{n} - 1} \frac{1}{2 ^{n}} \cdot \frac{1}{1 - \frac{r}{2 ^{n}}}

. As

n \to \infty

, this converges to the integral

\int_{0}^{1} \frac{1}{1 - x} d x

. Evaluating:

\int_{0}^{1} (1 - x)^{- 1/2} d x = [- 2 1 - x]_{0}^{1} = - 2 (0 - 1) = 2

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Single correct option

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