I have tried to use induction, but after I assume that P(n) is true, I can't go further to prove that P(n+1) is true as well. I also have tried to find an intermediate inequality, but I can't figure out which inequality I should start from.

Something that seemed to be useful was taking P(n) and multiplying it by $(1+\frac{1}{(n+1)^3})$, therefore I have come to this

$(1+ \frac{1}{1^3})(1+\frac{1}{2^3})...(1+\frac{1}{n^3})<3 | \times(1+\frac{1}{(n+1)^3})$

$(1+ \frac{1}{1^3})(1+\frac{1}{2^3})...(1+\frac{1}{n^3})(1+\frac{1}{(n+1)^3})<3(1+\frac{1}{(n+1)^3})$

but, as anyone could imagine, I came to contradiction because I've tried to prove that $3(1+\frac{1}{(n+1)^3})<3$ which is false.

Any help it would be useful.


1 Answer 1


Using the fact $1+x\le e^x$ for all real $x,$ we have $$\left(1+ \frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\cdots\left(1+\frac{1}{n^3}\right)\le \dfrac{9}{4}\exp\left(\sum_{k=3}^{n}\dfrac{1}{k^3}\right).$$ Now use the fact that $$\sum_{k=1}^{\infty}\dfrac{1}{k^3}\lt\dfrac{\pi^2}{7}.$$

  • $\begingroup$ Just curious ... Do you have an English language reference for Euler's proof that was cited in your link? He was a fairly sharp guy, that Leonard. $\endgroup$
    – Mark Viola
    Dec 4, 2020 at 21:34
  • $\begingroup$ @MarkViola: Indeed he is. This was first proved by Leonhard Euler in a 1772 paper, entitled Exercitationes Analyticae. I was unable to locate a translation of it. But, you can find a (different) derivation of that formula here. $\endgroup$
    – Bumblebee
    Dec 6, 2020 at 1:16

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