How to find this limit : $$\lim_{n \rightarrow \infty} \frac{1\cdot2+2\cdot3 +3\cdot4 +4\cdot5+\ldots+n(n+1)}{n^3}$$ As, if we look this limit problem viz. $\lim_{x \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}$ then we take the sum of numerator which is sum of first n natural numbers and we can write :
$$\lim_{x \rightarrow \infty} \frac{n(n+1)}{2n^2}$$ which gives after simplification :
$$ \frac{1}{2} $$ as other terms contain $\frac{1}{x}$ etc. and becomes zero.
$$\sum_{1\le r\le n}r(r+1)=\sum_{1\le r\le n}r^2+\sum_{1\le r\le n}r=\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}2$$
So, $$\lim_{n\to\infty}\frac{\sum_{1\le r\le n}r(r+1)}{n^3}$$ $$=\lim_{n\to\infty}\left(\frac{\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}2}{n^3}\right)$$
$$=\lim_{n\to\infty}\left(\frac{(1+\frac1n)(2+\frac1n)}6+\frac{(\frac1n+\frac1{n^2})}2\right)$$
$$=\frac{1\cdot2+0}6+\frac02=\frac13$$
Alternatively,
$$\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}2=\frac{n(n+1)}6\left(2n+1+3\right)=\frac{n(n+1)(n+2)}3$$
$$\implies\lim_{n\to\infty}\frac{\sum_{1\le r\le n}r(r+1)}{n^3}=\lim_{n\to\infty}\frac{n(n+1)(n+2)}{3n^3}=\frac13\lim_{n\to\infty}\left(1+\frac1n\right)\left(1+\frac2n\right)=\frac13$$
Squeeze principle can also help. Note that $$\lim_{n \to \infty}\frac{ 1\cdot 1 + 2\cdot 2 + 3 \cdot 3 + \cdots + n \cdot n}{n^{3}} \leq \lim_{n \to \infty}\frac{1\cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n\cdot(n+1)}{n^{3}}\leq \lim_{n\to\infty}\: \frac{2^{2}+3^{2}+4^{2}+\cdots +(n+1)^{2}}{n^{3}}$$
$$\Longrightarrow \frac{1}{3} \leq \lim_{n \to \infty}\frac{1\cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n\cdot(n+1)}{n^{3}} \leq \frac{1}{3}$$
By Riemann sum we have $$\frac{1}{n^3}\sum_{k=1}^nk(k+1)=\frac{1}{n}\sum_{k=1}^n\frac{k}{n}(\frac{k+1}{n})\sim_\infty \frac{1}{n}\sum_{k=1}^n\left(\frac{k}{n}\right)^2\to\int_0^1x^2dx=\frac{1}{3}$$
Or we can apply Stolz–Cesàro theorem to obtain $$\lim_{n \rightarrow \infty} \frac{1\cdot2+2\cdot3 +3\cdot4 +4\cdot5+\ldots+n(n+1)}{n^3}=\lim_{n \to \infty}\frac{(n+2)(n+1)}{(n+1)^3-n^3}=\lim_{n\to \infty}\frac{(n+2)(n+1)}{3n^2+3n+1}=\frac{1}{3} $$
