How do I evaluate $\lim_{n \rightarrow \infty} \frac{1\cdot2+2\cdot3 +3\cdot4 +4\cdot5+\ldots}{n^3}$ 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, 
$$\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}2=\frac{n(n+1)}6\left(2n+1+3\right)=\frac{n(n+1)(n+2)}3$$
$$\implies\lim_{n\to\infty}\frac{\sum_{1\le r\le n}r(r+1)}{n^3}=\lim_{n\to\infty}\frac{n(n+1)(n+2)}{3n^3}=\frac13\lim_{n\to\infty}\left(1+\frac1n\right)\left(1+\frac2n\right)=\frac13$$
