Question

# The value of $$\underset{n\rightarrow \infty}{\lim} \begin{Bmatrix}\dfrac{\sqrt{n+1}+\sqrt{n+2}+.....+\sqrt{2n-1}}{n^{\tfrac{3}{2}}}\end{Bmatrix}$$ is

A
23(221)
B
23(21)
C
23(2+1)
D
23(22+1)

Solution

## The correct option is A $$\dfrac{2}{3}(2\sqrt{2}-1)$$$$\underset{n\rightarrow \infty}{\lim} \begin{Bmatrix}\dfrac{\sqrt{n+1}+\sqrt{n+2}+.....+\sqrt{2n-1}}{n^{\tfrac{3}{2}}}\end{Bmatrix}$$$$=\underset{n\rightarrow \infty}{\lim} \begin{Bmatrix}\dfrac{\sqrt{n+1}+\sqrt{n+2}+.....+\sqrt{2n-1}}{\sqrt n}\end{Bmatrix}\dfrac{1}{n}$$$$=\underset{n\rightarrow \infty}{\lim} \begin{Bmatrix}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1+\dfrac{2}{n}}+.....+\sqrt{1+\dfrac{n-1}{n}}}{}\end{Bmatrix}\dfrac{1}{n}$$Now using limit as an integrals $$\displaystyle =\int_0^1\sqrt{1+x}dx=\dfrac{2}{3}[(1+x)\sqrt{1+x}]_0^1=\dfrac{2}{3}(2\sqrt2-1)$$Mathematics

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