Question

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

• A. $1/3$
• B. $1/5$
• C. $1/10$
• D. $1/2$

Answer 1

Answer: (A)

$1/3$

$\mu {\sum }_{r=0}^{n-1}{\frac{n^{1/2}}{(n+3(r\left)\right)}}^{3/2}$
$n\to \infty$
$\mu {\sum }_{r=0}^{n-1}{\frac{n^{1/2}}{(n^{3/2}+3(r\left)\right)}}^{3/2}$
$n\to \infty$
$\mu {\sum }_{r=0}^{n-1}\frac{1}{n}{\left(\frac{1}{1+\frac{3r}{n}}\right)}^{3/2}$
$n\to \infty$
${\int }_0^1\frac{1}{(1+3x{)}^{3/2}}dx=\left(\frac{-2}{3(1+3x{)}^{1/2}+c}\right){\int }_0^1$
$=\left([\frac{-2}{3(4{)}^{1/2}+c}-\frac{-2}{3}]\right)$
$-1/3+2/3=1/3.$