How would I show $$\lim_{n\to\infty}\sum_{k=1}^n(1+(k)/n)^3(1/n)=15/4?$$
My attempt is using the Riemann Sum technique. We know $(1+(k)/n)^2=f(\zeta_k)$ and $(1/n)=\Delta x$. So the definite integral goes from $a$ to $b$ where $b-a=1$.
I think I should find the formula for $$\sum_{k=1}^nk^3.$$ This would be $1+8+27+\cdots+n^3$... but then I get stuck.