Question

$n\stackrel{Lt}{\to }\infty [\frac{1}{n}+\frac{n^2}{(n+1{)}^3}+\frac{n^2}{(n+2{)}^3}+\dots .+\frac{1}{8n}]=$

• A. 2/5
• B. 3/5
• C. 3/8
• D. 11/7

Answer 1

The correct option is C 3/8

$=\underset{n\to \infty }{lim}\sum _{r=1}^{7n}\frac{n^2}{{(n+r)}^3}=\underset{n\to \infty }{lim}\sum _{r=1}^{n^2}\frac{n^2}{n^3{(1+\frac{r}{n})}^3}$
$=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^{2n}\frac{1}{{(1+\frac{r}{n})}^3}={\int }_0^7\frac{dx}{{(1+x)}^3}$
$={\left[\frac{{(1+x)}^{-2}}{-2}\right]}_0^7=\frac{3}{8}$