Alternative proof to an inequality regarding Apery constant

Question

Prove that:$$\sum_{k=1}^n \frac{1}{n^3}<1.23,$$ without Zeta-Riemann considerations.

I had absolutely no idea, and all my (naive) methods failed. Unfortunately, I am not allowed to use powerful tools in terms of real analysis (fast convergent series of that constant). To be specific, I am interested in a ,,classical solution'', with as little knowledge as possible.

A possible o.k. tool is using telescopic sums, and trying to get some range of that sum (this is what I tried). I know this sounds stupid, but these are the requirements. Also, by classical proof I mean a proof that is not assisted by computers, to compute a value of the partial sum, for huge terms.

P.S. Thanks for the downvotes, this will motivate me to improve myself.

Answer 1

• $\frac{1}{(n+1)^3}\lt \int\limits_n^{n+1} \frac1{x^3} \, dx$
• so $\sum\limits_{n=k}^\infty \frac{1}{(n+1)^3}\lt \int\limits_k^\infty\frac1{x^3} \, dx = \frac{1}{2k^2}$
• so $\sum\limits_{n=1}^\infty \frac{1}{n^3}\lt \sum\limits_{n=1}^k \frac{1}{n^3} +\frac{1}{2k^2}$

Here you can take $k=3$ and get $$\sum\limits_{n=1}^\infty \frac{1}{n^3}\lt \frac11 +\frac18 +\frac1{27}+\frac1{18} \approx 1.2176 <1.23 $$

Answer 2

Hint: $$\sum_{n=1}^{123} \frac{1}{n^3} + \sum_{n=123}^{\infty} \frac{1}{n^2} < 1.23$$