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)

$\displaystyle \lim _{n \rightarrow \infty} \frac{1}{2^n} \displaystyle\sum_{r=1}^{2^n} \frac{1}{\sqrt{1-\frac{r}{2^n}}}$
$\therefore \frac{1}{2^n} \rightarrow d x \Leftarrow \frac{r}{2^n}=x \left(\frac{r}{n^{\prime}}=x, \frac{1}{x}=d x\right) $
$ 2^n=n^{\prime}$
$ \displaystyle\lim _{n^{\prime} \rightarrow \infty} \frac{1}{n^{\prime}} \displaystyle\sum_{r=1}^{n^{\prime}-1} \frac{1}{\sqrt{1-\frac{r}{n^{\prime}}}}=\int\limits_0^1 \frac{1}{\sqrt{1-x}} d x $
$ \left.=-\frac{(1-x)^{1 / 2}}{1 / 2}\right]_0^1=-2[0-1]=2$