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.  

By Riemann sum we have
$$\frac{1}{n^3}\sum_{k=1}^nk(k+1)=\frac{1}{n}\sum_{k=1}^n\frac{k}{n}(\frac{k+1}{n})\sim_\infty \frac{1}{n}\sum_{k=1}^n\left(\frac{k}{n}\right)^2\to\int_0^1x^2dx=\frac{1}{3}$$
