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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.

Paul Rebenciuc
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2 Answers2

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

dtldarek
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  • Thanks, but I could not accept such a hint (unless you give more details). See my edited question. – Paul Rebenciuc Apr 18 '21 at 16:26
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    @PaulRebenciuc You write "I could not accept such a hint (unless you give more details)". This looks very much like you would like somebody to do your homework. Perhaps it is not, just looks this way, but without additional details it does look fishy, sorry. – dtldarek Apr 18 '21 at 16:38
  • It is not any homework, I just wanted to see alternative proofs. But I upvoted your hint, as a sign of gratitude. – Paul Rebenciuc Apr 18 '21 at 16:42
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  • $\frac{1}{(n+1)^3}\lt \int\limits_n^{n+1} \frac1{x^3} \, dx$
  • 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 $$\sum\limits_{n=1}^\infty \frac{1}{n^3}\lt \frac11 +\frac18 +\frac1{27}+\frac1{18} \approx 1.2176 <1.23 $$

Henry
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