Prove the following inequality: $\sum^n_{k=1} (k-1)^3 < \frac{n^4}{4}<\sum^n_{k=1} k^3$

Question

Prove the following inequality:

$$\sum^n_{k=1} (k-1)^3 < \frac{n^4}{4}<\sum^n_{k=1} k^3$$

Attempt:

By induction, we have:

Base Case: $$(1-1)^3< \frac{1^4}{4}< 1^3$$

Inductive step:

Assume it's true for $n=k$, then

Thesis:

$$\sum^k_{a=1}a^3 < \frac{(k+1)^4}{4}<\sum^{k+1}_{a=1} (a+1)^3$$

Demonstration:

If $\frac{k^4}{4}<\sum^{k}_{a=1} a^3,$ then $$\frac{(k-1)^4+4k^3}{4}<\sum^k_{a=1}(a-1)^3 < \frac{k^4}{4}$$ Adding $k^3$ on both sides, we have: $$\frac{(k-1)^4+4k^3}{4}<\sum^{k+1}_{a=1}(a+1)^3+ k^3 < \frac{ k^4+4k^3}{4}$$ $$(k-1)^4<k^4$$

I got here, but I don't know what to go on, or if it concludes anything. I would like help

Answer 1

Because $$\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}{4},$$ which we can prove by the telescoping summation.

Indeed, $$\sum_{k=1}^nk=\frac{1}{2}\sum_{k=1}^n((k+1)^2-k^2-1)=\frac{1}{2}((n+1)^2-1-n)=\frac{n(n+1)}{2};$$ $$\sum_{k=1}^nk^2=\frac{1}{3}\sum_{k=1}^n((k+1)^3-k^3-3k-1)=$$ $$=\frac{1}{3}\left((n+1)^3-1-\frac{3n(n+1)}{2}-n\right)=\frac{n(n+1)(2n+1)}{6}$$ and $$\sum_{k=1}^nk^3=\frac{1}{4}\sum_{k=1}^n((k+1)^4-k^4-6k^2-4k-1)=$$ $$=\frac{1}{4}\left((n+1)^4-1-\frac{6n(n+1)(2n+1)}{6}-\frac{4n(n+1)}{2}-n\right)=\frac{n^2(n+1)^2}{4}.$$ Thus, it's enough to prove that: $$\frac{(n-1)^2n^2}{4}<\frac{n^4}{4}<\frac{n^2(n+1)^2}{4},$$ which is obvious.

Answer 2

If you know calculus, then observe that $$\frac{n^4}{4} = \int_0^n x^3\,dx = \sum_{k=1}^{n}\int_{k-1}^k x^3\,dx.$$ Since $x^3$ is an increasing function, it holds $$(k-1)^3 < \int_{k-1}^kx^3\,dx < k^3,$$ hence $$\sum_{k=1}^n(k-1)^3 < \frac{n^4}{4} < \sum_{k=1}^n k^3.$$