The value of $\underset{n\to \mathrm{\infty }}{lim} \frac{1\cdot 2+2\cdot 3+3\cdot 4+\dots +n(n+1)}{n^3}$ ,is 

Solution:

We have, 

$\begin{array}{l}\underset{n\to \mathrm{\infty }}{lim} \frac{1\cdot 2+2\cdot 3+3\cdot 4+\dots +n(n+1)}{n^3}\\ =\underset{n\to \mathrm{\infty }}{lim} \frac{\sum _{r=1}^n r(r+1)}{n^3}=\underset{n\to \mathrm{\infty }}{lim} \frac{\sum _{r=1}^n r^2+\sum _{r=1}^n r}{n^3}\\ =\underset{n\to \mathrm{\infty }}{lim} \frac{\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}}{n^3}\\ =\underset{n\to \mathrm{\infty }}{lim} \frac{n(n+1)(n+2)}{3n^3}=\frac{1}{3}\end{array}$