lim_(n→∞) ((((√(n+1))+(√(n+2))+(√(n+3))+...+(√(2n−1)))/n^(3/2) ) ) =


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Question Number 116317 by bemath last updated on 03/Oct/20

$\lim_{n \to \infty} (\frac{\sqrt{n + 1} + \sqrt{n + 2} + \sqrt{n + 3} + . . . + \sqrt{2 n - 1}}{n^{\frac{3}{2}}}) =$
	Answered by Bird last updated on 03/Oct/20
	$U_n =(1/n^(3/2) )Σ_(k=1) ^(n−1) (√(n+k)) =(1/n^(3/2) ).(√n)Σ_(k=1) ^(n−1) (√(1+(k/n))) =(1/n^((3/2)−(1/2)) )Σ_(k=1) ^(n−1) (√(1+(k/n))) =(1/n)Σ_(k=1) ^(n−1) (√(1+(k/n)))→∫_0 ^1 (√(1+x))dx ∫_0 ^1 (√(1+x))dx =_((√(1+x))=t) ∫_1 ^(√2) t(2t)dt =2 ∫_1 ^(√2) t^(2 ) dt =(2/3)[t^3 ]_1 ^(√2) =(2/3){3(√2)−1} =2(√2)−(2/3)$
	$U_{n} = \frac{1}{n^{\frac{3}{2}}} \sum_{k = 1}^{n - 1} \sqrt{n + k}$ $= \frac{1}{n^{\frac{3}{2}}} . \sqrt{n} \sum_{k = 1}^{n - 1} \sqrt{1 + \frac{k}{n}}$ $= \frac{1}{n^{\frac{3}{2} - \frac{1}{2}}} \sum_{k = 1}^{n - 1} \sqrt{1 + \frac{k}{n}}$ $= \frac{1}{n} \sum_{k = 1}^{n - 1} \sqrt{1 + \frac{k}{n}} \to \int_{0}^{1} \sqrt{1 + x} d x$ $\int_{0}^{1} \sqrt{1 + x} d x =_{\sqrt{1 + x} = t} \int_{1}^{\sqrt{2}} t (2 t) d t$ $= 2 \int_{1}^{\sqrt{2}} t^{2} d t = \frac{2}{3} {[t^{3}]}_{1}^{\sqrt{2}}$ $= \frac{2}{3} {3 \sqrt{2} - 1} = 2 \sqrt{2} - \frac{2}{3}$
	Commented by john santu last updated on 03/Oct/20

	$t y p o = \frac{2}{3} (2 \sqrt{2} - 1)$
	Commented by Bird last updated on 03/Oct/20

	$y e s t h a n k s j o h n$
	Answered by Dwaipayan Shikari last updated on 03/Oct/20

	$\frac{1}{n} \lim_{n \to \infty} (\sqrt{1 + \frac{1}{n}} + \sqrt{1 + \frac{2}{n}} + . . . . .)$ $\lim_{n \to \infty} \frac{1}{n} \underset{k = 1}{\sum^{n}} (\sqrt{1 + \frac{k}{n}})$ $\int_{0}^{1} \sqrt{1 + x} dx$ $= \frac{2}{3} {[{(1 + x)}^{\frac{3}{2}}]}_{0}^{1} = \frac{2}{3} (2 \sqrt{2} - 1)$
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