lim-n-n-1-n-2-n-3-2n-1-n-3-2-

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} + \dots + \sqrt{2 n - 1}}{n^{\frac{3}{2}}}) =

Answered by Bird last updated on 03/Oct/20

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}} + \dots . .)

\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)