Question

${lim}_{n\to \infty }(\frac{1}{n}+\frac{n^2}{(n+1{)}^3}+\frac{n^2}{(n+2{)}^3}+\dots +\frac{1}{8n})$ is equal to

Answer 1

$\begin{array}{c}\frac{1}{n}+\frac{n^2}{{\left(n+1\right)}^3}+\frac{n^2}{{\left(n+2\right)}^3}+...........+\frac{n^2}{{\left(n+n\right)}^3}\\ =\frac{n^2}{n^3}+\frac{n^2}{{\left(n+1\right)}^3}+\frac{n^2}{{\left(n+2\right)}^3}+.........+\frac{n^2}{{\left(n+n\right)}^3}\\ =\frac{1}{n}\left[\frac{n^3}{n^3}+\frac{n^3}{{\left(n+1\right)}^3}+\frac{n^3}{{\left(n+2\right)}^3}+...........+\frac{n^3}{{\left(n+n\right)}^3}\right]\\ =\frac{1}{n}\left[\frac{1}{{\left(\frac{n}{n}\right)}^3}+\frac{1}{{\left(\frac{n}{n}+\frac{1}{n}\right)}^3}+\frac{1}{{\left(\frac{n}{n}+\frac{2}{n}\right)}^3}+...........+\frac{1}{{\left(\frac{n}{n}+\frac{n}{n}\right)}^3}\right]\\ ={\int }_0^1\frac{dx}{{\left(1+x\right)}^3}\\ =\frac{{\left[{\left(1+x\right)}^{-2}\right]}_0^1}{-2}\\ =\frac{-1}{2}{\left[\frac{1}{{\left(1+x\right)}^2}\right]}_0^1\\ =-\frac{1}{2}\left[\frac{1}{4}-1\right]\\ =\frac{1}{2}\left[1-\frac{1}{4}\right]\\ =\frac{1}{2}\times \frac{3}{4}\\ =\frac{3}{8}\end{array}$