Question

$\underset{n\to \infty }{Lt}[\frac{n^{1/2}}{(n+3{)}^{3/2}}+\frac{n^{1/2}}{(n+6{)}^{3/2}}+\frac{n^{1/2}}{(n+9{)}^{3/2}}+...+\frac{1}{8n}]$ is equal to :

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{3}$
• D. None of these

Answer 1

The correct option is C $\frac{1}{3}$

$\underset{n\to \infty }{Lim}[\frac{n^{\frac{1}{2}}}{{(n+3)}^{\frac{3}{2}}}+\frac{n^{\frac{1}{2}}}{{(n+6)}^{\frac{3}{2}}}+....\frac{1}{8n}]$
$\underset{n\to \infty }{Lim}\frac{1}{n}{\sum }_{r=1}^n\frac{1}{{(1+\frac{3r}{n})}^{3/2}}={\int }_0^1\frac{1}{(1+3x{)}^{3/2}}$
let $1+3x=t\Rightarrow 3dx=dt$
$\Rightarrow {\int }_0^1\frac{1}{(1+3x{)}^{3/2}}=\frac{1}{3}{\int }_1^4\frac{dt}{t^{3/2}}$
$=\frac{1}{3}\left[\frac{t^{\frac{3}{2}+1}}{-\frac{3}{2}+1}\right]=\frac{1}{3}\left[\frac{{\left(4\right)}^{\frac{-1}{2}}-{\left(1\right)}^{\frac{-1}{2}}}{-\frac{1}{2}}\right]$
$=\frac{-2}{3}[\frac{1}{2}-1]=-\frac{2}{3}\times \frac{-1}{2}=\frac{1}{3}$