# How to show $\lim_{n\to\infty}\sum_{k=1}^n(1+(k)/n)^3(1/n)=15/4$?

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.

• Ask Faulhaber... – Yves Daoust Dec 3 '18 at 14:30
• As a Riemannian sum, $\int_0^1(1+x)^3dx=15/4$. – Yves Daoust Dec 3 '18 at 14:31
• This is odd, you are asked to find the limit of some sums, thus you turn to integrals (good idea), but then, instead of evaluating the relevant integral (a trivial task), you suddenly make a U-turn to come back to sums... where you declare you are stuck. – Did Dec 3 '18 at 14:52

$$S_n:=\sum_{k=1}^n k^3$$ must be a quartic polynomial in $$n$$ because $$S_{n}-S_{n-1}=n^3$$ is a cubic polynomial.
$$S_n-S_{n-1}=an^4+bn^3+cn^2+\cdots-a(n-1)^4-b(n-1)^3-c(n-1)^2-\cdots \\=4an^3+bn^3-bn^3+\cdots$$ and lower order terms. You conclude that
$$S_n\sim\frac{n^4}4$$ and this is enough for you to evaluate the limit.