Question

$\underset{n\to \infty }{\text{lim}}(\frac{1}{n}+\frac{n^2}{(n+1{)}^3}+\frac{n^2}{(n+2{)}^3}+...+\frac{1}{8n})$ is equal to

• A. $\frac{3}{8}$
• B. $\frac{1}{4}$
• C. $\frac{1}{8}$
• D. $\frac{7}{8}$

Answer 1

The correct option is A

$\frac{3}{8}$

$\underset{n\to \infty }{\text{lim}}\left[\right(\frac{1}{n}+\frac{n^2}{(n+1{)}^3}+\frac{n^2}{(n+2{)}^3}+...+\frac{1}{8n}\left)\right]$

=$\underset{n\to \infty }{\text{lim}}[\frac{n^2}{(n+0{)}^3}+\frac{n^2}{(n+1{)}^3}+\frac{n^2}{(n+2{)}^3}+....+\frac{n^2}{(n+n{)}^3}]$

=$\underset{n\to \infty }{\text{lim}}{\sum }_{r=0}^n\frac{1}{n}\frac{1}{(1+\frac{r}{n}{)}^3}$

$={\int }_0^1\frac{dx}{(1+x{)}^3}=[-\frac{1}{2(1+x{)}^2}{]}_0^1$

$=-\frac{1}{2}(\frac{1}{4}-1)=\frac{3}{8}$