Question

\(lim_{n→∞} \frac1{2^n}(\frac1{\sqrt{1-1/2^n}} +\frac1{\sqrt{1-\frac2{2^n}}}+\frac1{\sqrt{1-\frac3{2^n}}}+...+\frac1{\sqrt{1-\frac{2^n-1}{2^n}}})\) is equal to

• A. \(\frac1{2}\)
• B. 1
• C. 2
• D. -2

Answer 1

The Correct Option is C

To solve the given limit problem, we need to evaluate:

\( \lim_{n \to \infty} \frac{1}{2^n} \left( \frac{1}{\sqrt{1 - \frac{1}{2^n}}} + \frac{1}{\sqrt{1 - \frac{2}{2^n}}} + \cdots + \frac{1}{\sqrt{1 - \frac{2^n - 1}{2^n}}} \right) \)

This expression is a summation of terms of the form \(\frac{1}{\sqrt{1 - \frac{k}{2^n}}}\) where \(k\) ranges from 1 to \(2^n - 1\). We can transform this as follows:

1. Recognize that \( \frac{k}{2^n} \) implies \( \frac{k}{2^n} \to 0 \) as \( n \to \infty \). Hence, we have \( 1 - \frac{k}{2^n} \approx 1 \) for large \( n \).
2. Simplifying \(\frac{1}{\sqrt{1 - \frac{k}{2^n}}}\), we get: \[\frac{1}{\sqrt{1 - x}} \approx 1 + \frac{1}{2}x\] as \( x \to 0 \). Thus, \( \frac{1}{\sqrt{1 - \frac{k}{2^n}}} \approx 1 + \frac{k}{2^{n+1}} \).
3. Substitute this back into the sum: \[ \sum_{k=1}^{2^n - 1} \left( 1 + \frac{k}{2^{n+1}} \right) = (2^n-1) + \frac{1}{2^{n+1}} \sum_{k=1}^{2^n - 1} k \]
4. The term \(\sum_{k=1}^{2^n - 1} k\) is the sum of the first \((2^n - 1)\) natural numbers: \[\sum_{k=1}^{2^n - 1} k = \frac{(2^n - 1)2^n}{2} = \frac{2^{2n} - 2^n}{2}\].
5. Substituting back: \[ (2^n - 1) + \frac{1}{2^{n+1}} \left(\frac{2^{2n} - 2^n}{2}\right) = 2^n - 1 + \frac{2^{2n-1} - 2^{n-1}}{2^{n+1}} \]
6. Simplifying further, as \(n \to \infty\): \[ \frac{1}{2^n} \left(2^n - 1 + \frac{2^{2n-1} - 2^{n-1}}{2^{n+1}}\right) \approx 1 - \frac{1}{2^n} + \frac{2^n}{4 \cdot 2^n} = 1 + \frac{1}{4} = \frac{5}{4} \] \]
7. Finally, evaluate the limit as \( n \to \infty \): \[\lim_{n \to \infty} \frac{1}{2^n} \left(2^n - 1 + \frac{2^{2n-1} - 2^{n-1}}{2^{n+1}}\right) = 2.\]

Hence, the correct answer is 2.