<s>[INST] 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.  
 [/INST] Or we can apply Stolz–Cesàro theorem to obtain $$\lim_{n \rightarrow \infty} \frac{1\cdot2+2\cdot3 +3\cdot4 +4\cdot5+\ldots+n(n+1)}{n^3}=\lim_{n \to \infty}\frac{(n+2)(n+1)}{(n+1)^3-n^3}=\lim_{n\to \infty}\frac{(n+2)(n+1)}{3n^2+3n+1}=\frac{1}{3} $$
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