1

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

Assandra Lakal
  • 185
  • 1
    To prove the induction you don't need to prove $\sum^k_{a=1}a^3 < \frac{(k+1)^4}{4}<\sum^{k+1}{a=1} (a+1)^3$, but $\sum^{k+1}{a=1}(a-1)^3 < \frac{(k+1)^4}{4}<\sum^{k+1}_{a=1} a^3$ – jjagmath May 28 '23 at 20:00
  • 1
    Also, it's confusing having $k$ in your question statement and in "Assume it's true for $n=k$". One way to remedy this would be to instead, say:

    $$$$

    Assume it's true for $n=m$, then $$$$ Thesis:

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

     – Adam Rubinson May 28 '23 at 20:27
  • 1
    Also: https://math.stackexchange.com/q/2630761/42969 and https://math.stackexchange.com/q/318357/42969 – all found with Approach0 – Martin R May 28 '23 at 20:55
  • @Martin R You continue to destroy our forum. This problem is not duplicate. Topic starter does not want a solution by induction only. In my solution I did not use induction. – Michael Rozenberg May 28 '23 at 21:26
  • 2
    @MichaelRozenberg: Is is a duplicate in my opinion. I linked to four Q&As where the problem (for general exponent $p$) is solved with different methods, with and without induction. – Martin R May 28 '23 at 21:48
  • I agree with @MartinR that the first question linked is a better duplicate of this question. – Clayton May 28 '23 at 23:43
  • @Clayton: Please vote to close as a duplicate so that others can chime it. I cannot vote again because my initial duplicate vote had been reverted by another gold-badge holder. – Martin R May 29 '23 at 08:28
  • @MichaelRozenberg The question does not say anything about the proof method. They got stuck with induction, but that is not the same as not wanting a solution by induction only! Moreover, considering they used the "induction" tag, it is reasonable to assume they do want a proof by induction. – user1729 May 29 '23 at 09:22

2 Answers2

1

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

BBBBBB
  • 1,227
0

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.

Michael Rozenberg
  • 194,933