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.  

Squeeze principle can also help. Note that $$\lim_{n \to \infty}\frac{ 1\cdot 1 + 2\cdot 2 + 3 \cdot 3 + \cdots + n \cdot n}{n^{3}} \leq \lim_{n \to \infty}\frac{1\cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n\cdot(n+1)}{n^{3}}\leq \lim_{n\to\infty}\: \frac{2^{2}+3^{2}+4^{2}+\cdots +(n+1)^{2}}{n^{3}}$$
$$\Longrightarrow \frac{1}{3} \leq \lim_{n \to \infty}\frac{1\cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n\cdot(n+1)}{n^{3}} \leq \frac{1}{3}$$
