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Cubic Sums Bound a Quartic Term

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    Vu Hung
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Problem Statement

Prove by induction that for all integers n≥2n \geq 2:

13+23+⋯+(n−1)3<n44<13+23+⋯+n31^3 + 2^3 + \cdots + (n-1)^3 < \frac{n^4}{4} < 1^3 + 2^3 + \cdots + n^3

Hints

This is a "sandwich" inequality with two parts:

  • LHS inequality: ∑k=1n−1k3<n44\sum_{k=1}^{n-1} k^3 < \frac{n^4}{4} (needs induction)
  • RHS inequality: n44<∑k=1nk3\frac{n^4}{4} < \sum_{k=1}^{n} k^3 (can be proven directly)

Recall: ∑k=1nk3=[n(n+1)2]2=n2(n+1)24\sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2 = \frac{n^2(n+1)^2}{4}

For the LHS induction step, you'll need to show k44+k3<(k+1)44\frac{k^4}{4} + k^3 < \frac{(k+1)^4}{4}, which simplifies to 0<6k2+4k+10 < 6k^2 + 4k + 1.

Solutions

Part 1: RHS Inequality (Direct Proof)

Show n44<∑k=1nk3\frac{n^4}{4} < \sum_{k=1}^{n} k^3 for n≥2n \geq 2.

Using the sum formula:

∑k=1nk3=n2(n+1)24=n4+2n3+n24\sum_{k=1}^{n} k^3 = \frac{n^2(n+1)^2}{4} = \frac{n^4 + 2n^3 + n^2}{4}

We need: n44<n4+2n3+n24\frac{n^4}{4} < \frac{n^4 + 2n^3 + n^2}{4}

This simplifies to: 0<2n3+n2=n2(2n+1)0 < 2n^3 + n^2 = n^2(2n + 1)

Since n≥2>0n \geq 2 > 0, this is clearly true. RHS proven. \checkmark

Part 2: LHS Inequality (Induction)

Prove ∑k=1n−1k3<n44\sum_{k=1}^{n-1} k^3 < \frac{n^4}{4} for n≥2n \geq 2.

Base case (n=2n=2):

∑k=11k3=13=1<244=4✓\sum_{k=1}^{1} k^3 = 1^3 = 1 < \frac{2^4}{4} = 4 \quad \checkmark

Inductive hypothesis: Assume ∑k=1m−1k3<m44\sum_{k=1}^{m-1} k^3 < \frac{m^4}{4} for some m≥2m \geq 2.

Inductive step: Prove ∑k=1mk3<(m+1)44\sum_{k=1}^{m} k^3 < \frac{(m+1)^4}{4}.

∑k=1mk3=∑k=1m−1k3+m3<m44+m3(by IH)=m4+4m34\begin{aligned} \sum_{k=1}^{m} k^3 &= \sum_{k=1}^{m-1} k^3 + m^3 \\ &< \frac{m^4}{4} + m^3 \quad \text{(by IH)} \\ &= \frac{m^4 + 4m^3}{4} \end{aligned}

Need to show: m4+4m34<(m+1)44\frac{m^4 + 4m^3}{4} < \frac{(m+1)^4}{4}

Equivalently: m4+4m3<(m+1)4m^4 + 4m^3 < (m+1)^4

Expand RHS:

(m+1)4=m4+4m3+6m2+4m+1(m+1)^4 = m^4 + 4m^3 + 6m^2 + 4m + 1

So we need: m4+4m3<m4+4m3+6m2+4m+1m^4 + 4m^3 < m^4 + 4m^3 + 6m^2 + 4m + 1

Simplifies to: 0<6m2+4m+10 < 6m^2 + 4m + 1

This is true for all m≥2m \geq 2. \checkmark

By induction, LHS proven. Combining both parts, the full sandwich inequality holds. ■\blacksquare

Takeaways

  • Reconstruct the full proof from the hint and the solution outline, and justify every transformation explicitly.
  • Check edge cases and verify where each assumption is used in the argument.

Further Readings

If you found this proof interesting, be sure to check out these relevant HSC booklets to sharpen your reasoning skills:

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