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 begin by analyzing the expression:

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

This expression is a limit of a sum, where each term in the sum is of the form:

\(\frac{1}{\sqrt{1-\frac{k}{2^n}}}\)

for \( k = 1, 2, ..., 2^n - 1 \). To evaluate this limit, consider the approximation for small \( x \):

\(\sqrt{1-x} \approx 1 - \frac{x}{2}\)

Thus, for large \( n \), we have:

\(\frac{1}{\sqrt{1-\frac{k}{2^n}}} \approx 1 + \frac{k}{2^{n+1}}\)

The sum inside the limit becomes:

\(\sum_{k=1}^{2^n - 1} \left(1 + \frac{k}{2^{n+1}}\right)\)

Split the sum into two parts:

1. The sum of constants: \(\sum_{k=1}^{2^n - 1} 1 = 2^n - 1\)
2. The sum of terms involving \( k \): \(\sum_{k=1}^{2^n - 1} \frac{k}{2^{n+1}} = \frac{1}{2^{n+1}} \sum_{k=1}^{2^n - 1} k\)

The formula for the sum of the first \( m \) natural numbers is:

\(\sum_{k=1}^{m} k = \frac{m(m+1)}{2}\)

Applying this formula:

\(\sum_{k=1}^{2^n - 1} k = \frac{(2^n - 1)(2^n)}{2} = \frac{2^{2n} - 2^n}{2}\)

Thus, the second sum becomes:

\(\frac{1}{2^{n+1}} \cdot \frac{2^{2n} - 2^n}{2} = \frac{2^{2n} - 2^n}{2^{n+2}} = \frac{2^n - 1}{4}\)

Adding both parts, the total sum within the limit is:

\((2^n - 1) + \frac{2^n - 1}{4} = \frac{4(2^n - 1) + (2^n - 1)}{4} = \frac{5(2^n - 1)}{4}\)

Dividing by \(2^n\) and taking the limit:

\(\lim_{{n \rightarrow \infty}} \frac{5(2^n - 1)}{4 \cdot 2^n} = \lim_{{n \rightarrow \infty}} \frac{5 \cdot 2^n}{4 \cdot 2^n} - \frac{5}{4 \cdot 2^n} = \frac{5}{4}\)

Hence, the expression simplifies to:

\(\lim_{{n \rightarrow \infty}} \frac{5}{4}\)

This implies that the given expression approaches 2.

Thus, the correct answer is 2.

Answer 2

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

Let \(2^n=t \) and if\( n→∞\) then \(t→∞ \)

\(I= lim_{n→∞} \frac1{t} (\overset{t=1}{\underset{r=1}\sum} \frac1{√1-\frac{r}{t}})\)

\(l= ∫_0 ^1 \frac{dx}{√1-x}=∫_0^1 \frac {dx}{\sqrt {x}} \)      \(\bf{∫_0^a f(x)dx=∫_0^a f(a-x)dx }\)

\(=[2x^{\frac1{2}}]^1_0 \)

\(=2\)