Not everything get cancelled

Relevant wiki: Sum of n, n², or n³

Note that

$\begin{aligned} \sum_{k=1}^n k & = \frac {n(n+1)}2 \\ \sum_{k=1}^n k^2 & = \frac {n(n+1)(2n+1)}6 \\ \sum_{k=1}^n k^3 & = \left(\frac {n(n+1)}2\right)^2 \\ \sum_{k=1}^n k^4 & = \frac {n(n+1)(2n+1)(3n^2+3n-1}{30} \end{aligned}$

Therefore, $\begin{cases} A = \dfrac {\sum_{k=1}^n k}{\sum_{k=1}^n k^3} = \dfrac 2{n(n+1)} \\ B = \dfrac {\sum_{k=1}^n k^2}{\sum_{k=1}^n k^4} = \dfrac 5{3n^2+3n-1} \end{cases}$

$\implies \dfrac AB = \dfrac {2(3n^2+3n-1)}{5(n^2+n)} = \dfrac {6\left(n^2+n-\frac 13\right)}{5(n^2+n)} = \dfrac 65 - \dfrac 2{5(n^2+n)} > 1$ for $n > 1$ , therefore $\boxed{A>B}$ .