Let $I=\underset{n\to \infty }{\lim }\frac{1}{2^n}\left(\frac{1}{\sqrt{1-\frac{1}{2^n}}}+\frac{1}{\sqrt{1-\frac{2}{2^n}}}+\frac{1}{\sqrt{1-\frac{3}{2^n}}}+\dots .+\frac{1}{\sqrt{1-\frac{2^n-1}{2^n}}}\right)$

Let $2^n=t$ and if $n\to \infty$ then $t\to \infty$

$I=\underset{n\to \infty }{\lim }\frac{1}{t}\left(\sum _{r=1}^{t-1}\frac{1}{\sqrt{1-\frac{r}{t}}}\right)$

$I={\int }_0^1\frac{dx}{\sqrt{1-x}}$ (Using definite integral as limit of sum)

Applying ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx$, we get

$I={\int }_0^1\frac{dx}{\sqrt{x}}$

$={\left[2\sqrt{x}\right]}_0^1=2$