Question

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

• A. $\frac{1}{2}$
• B. 1
• C. 2
• D. $-2$

Answer 1

Answer: (C)

2

Answer 2

n→∞lim​2n1​r=1∑2n​1−2nr​​1​
∴2n1​→dx⇐2nr​=x(n′r​=x,x1​=dx)
2n=n′
n′→∞lim​n′1​r=1∑n′−1​1−n′r​​1​=0∫1​1−x​1​dx
=−1/2(1−x)1/2​]01​=−2[0−1]=2