Mathematics

# $\displaystyle \lim_{n\to \infty} \left[ \frac{n^{1/2}}{n^{3/2}} +\frac{n^{1/2}}{(n+3)^{3/2}}+\frac{n^{1/2}}{(n+6)^{3/2}}+......\frac{n^{1/2}}{ \left \{n+3(n-1)\right \}^{3/2} } \right]=$

$1/3$

##### SOLUTION

$\mu \sum_{r=0}^{n-1}\dfrac{n^{1/2}}{(n+3(r))}^{3/2}$
$n\rightarrow \infty$
$\mu \sum_{r=0}^{n-1}\dfrac{n^{1/2}}{(n^{3/2}+3(r))}^{3/2}$
$n\rightarrow \infty$
$\mu \sum_{r=0}^{n-1}\dfrac{1}{n}\left ( \dfrac{1}{1+\dfrac{3r}{n}} \right )^{3/2}$
$n\rightarrow \infty$
$\int_{0}^{1}\dfrac{1}{(1+3x)^{3/2}}dx=\left ( \dfrac{-2}{3(1+3x)^{1/2}+c} \right )\int_{0}^{1}$
$=\left ( \left [ \dfrac{-2}{3(4)^{1/2}+c}-\dfrac{-2}{3} \right ] \right )$
$-1/3+2/3=1/3.$

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 126

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