Question

Evaluate: $\underset{n\to \infty }{\lim }\frac{1.2+2.3+3.4+...+n\left(n+1\right)}{n^3}$

Answer 1

$$\begin{gathered} \underset{n\to \infty }{\lim }\frac{1.2+2.3+3.4+...+n\left(n+1\right)}{n^3} \\ =\underset{n\to \infty }{\lim }\frac{{\displaystyle \sum _{k=1}^n}k\left(k+1\right)}{n^3} \\ =\underset{n\to \infty }{\lim }\frac{{\displaystyle \sum _{k=1}^n{k}^2+\sum _{k=1}^{n}k}}{n^3} \\ =\underset{n\to \infty }{\lim }\frac{{\displaystyle \frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{n\left(n+1\right)}{2}}}{n^3} \end{gathered}$$
$$\begin{gathered} =\underset{n\to \infty }{\lim }\frac{n\left(n+1\right)\left(2n+1+3\right)}{6n^3} \\ =\underset{n\to \infty }{\lim }\frac{2n\left(n+1\right)\left(n+2\right)}{6n^3} \\ =\frac{1}{3}\underset{n\to \infty }{\lim }\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right) \\ =\frac{1}{3}\times \left(1+0\right)\times \left(1+0\right)\left(\underset{n\to \infty }{\lim }\frac{1}{n}=0\right) \\ =\frac{1}{3} \end{gathered}$$