Q:

Show that for all $n ≥ 2$ it is true: $1^3+2^3+\cdots+(n-1)^3<\dfrac{n^4}{4}$

How can I prove that?
$1^3+2^3+\cdots+(n-1)^3<\frac{n^4}{4}$ 

A:

If you compare with the integral of $x\mapsto x^3$ (increasing function):
$$\sum_{k=1}^{n-1}k^3\leq \sum_{k=1}^{n-1}\int_{k}^{k+1}x^3dx=\int_1^nx^3dx=\frac{n^4-1}4<\frac {n^4} 4$$

