lim _{n rightarrow infty} frac{1}{2^{n}}left( | vedclass

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(C) Let the given limit be $I = \lim _{n \rightarrow \infty} \frac{1}{2^{n}} \sum_{r=1}^{2^{n}-1} \frac{1}{\sqrt{1-\frac{r}{2^{n}}}}$.Substitute $2^{n

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Let the given limit be $I = \lim _{n \rightarrow \infty} \frac{1}{2^{n}} \sum_{r=1}^{2^{n}-1} \frac{1}{\sqrt{1-\frac{r}{2^{n}}}}$.Substitute $2^{n} = t$. As $n \rightarrow \infty$,$t \rightarrow \infty$.The expression becomes $I = \lim _{t \rightarrow \infty} \frac{1}{t} \sum_{r=1}^{t-1} \frac{1}{\sqrt{1-\frac{r}{t}}}$.This is a Riemann sum of the form $\int_{0}^{1} f(x) dx = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f(\frac{r}{n})$.Here,$f(x) = \frac{1}{\sqrt{1-x}}$.Thus,$I = \int_{0}^{1} \frac{1}{\sqrt{1-x}} dx$.Let $u = 1-x$,then $du = -dx$. When $x=0, u=1$ and when $x=1, u=0$.$I = \int_{1}^{0} \frac{1}{\sqrt{u}} (-du) = \int_{0}^{1} u^{-1/2} du$.$I = [2u^{1/2}]_{0}^{1} = 2(1) - 2(0) = 2$.

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

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