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

The Correct Option is C

$\underset{n\to \infty }{\lim }\frac{1}{2^n}\sum _{r=1}^{2^n}\frac{1}{\sqrt{1-\frac{r}{2^n}}}$
$\therefore \frac{1}{2^n}\to dx\Leftarrow \frac{r}{2^n}=x\left(\frac{r}{n^{\prime }}=x,\frac{1}{x}=dx\right)$
$2^n=n^{\prime }$
$\underset{n^{\prime }\to \infty }{\lim }\frac{1}{n^{\prime }}\sum _{r=1}^{n^{\prime }-1}\frac{1}{\sqrt{1-\frac{r}{n^{\prime }}}}=\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{1-x}}dx$
${=-\frac{(1-x{)}^{1/2}}{1/2}]}_0^1=-2[0-1]=2$