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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.

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  • $\begingroup$ Why 1.23? What that bound? $\endgroup$ Apr 18, 2021 at 16:06
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    $\begingroup$ Can you hack something from the Apéry's Constant Wiki Page to prove this? $\endgroup$ Apr 18, 2021 at 16:08
  • $\begingroup$ @MikePierce I am afraid I cannot use stronger machinery than zeta-Riemann, or similar to this, I need a classic proof. $\endgroup$ Apr 18, 2021 at 16:10
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    $\begingroup$ Please explain what you mean by $\Delta$ peer pressure ? $\endgroup$
    – Jean Marie
    Apr 18, 2021 at 16:32
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    $\begingroup$ Sorry, nothing, forget about it. $\endgroup$ Apr 18, 2021 at 16:34

2 Answers 2

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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 $$\sum\limits_{n=1}^\infty \frac{1}{n^3}\lt \frac11 +\frac18 +\frac1{27}+\frac1{18} \approx 1.2176 <1.23 $$

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  • $\begingroup$ This is a great answer, thank you for your time! $\endgroup$ Apr 18, 2021 at 16:45
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Hint: $$\sum_{n=1}^{123} \frac{1}{n^3} + \sum_{n=123}^{\infty} \frac{1}{n^2} < 1.23$$

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  • $\begingroup$ Thanks, but I could not accept such a hint (unless you give more details). See my edited question. $\endgroup$ Apr 18, 2021 at 16:26
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    $\begingroup$ @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. $\endgroup$
    – dtldarek
    Apr 18, 2021 at 16:38
  • $\begingroup$ It is not any homework, I just wanted to see alternative proofs. But I upvoted your hint, as a sign of gratitude. $\endgroup$ Apr 18, 2021 at 16:42

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