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Prove that

$$\sum_{n=1}^{\infty} \frac{\mathcal{H}_n}{n} \left( \zeta(2) - 1-\frac{1}{2^2} \cdots -\frac{1}{n^2} \right)\left ( \zeta(3) - 1-\frac{1}{2^3}-\cdots - \frac{1}{n^3} \right) = \frac{11}{4} \zeta(3) \zeta(4) -2\zeta(2) \zeta(5)$$

where $\mathcal{H}_n$ denotes the $n$ - th harmonic number and $\zeta$ the Riemann function.

This sum has been proposed $5$ years ago but I cannot figure out a solution. I don't even know where to start.

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    $\begingroup$ If it was proposed $5$ years ago, please give the source! $\endgroup$ Commented Nov 9, 2022 at 19:06
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    $\begingroup$ I agree that you should tell us where the problem comes from, but since this is clearly not a homework problem, and it seems you are posting this more out of curiosity, I don't think you deserve downvotes here. Have a (+1) to counteract the (-1). $\endgroup$ Commented Nov 9, 2022 at 19:20
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    $\begingroup$ The problem was proposed by Cornel Valean ... $\endgroup$
    – Tolaso
    Commented Nov 10, 2022 at 7:57
  • $\begingroup$ @Tolaso I agree with you! Thank you for the bounty! :-) $\endgroup$ Commented Nov 10, 2022 at 8:06

1 Answer 1

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This is one of the most beautiful and difficult problems in (Almost) Impossible Integrals, Sums, and Series, where a miraculous solution (that's because it circumvents the necessity of knowing and using the values of all resulting advanced harmonic series of weight $7$ appearing during the calculations) is given, by only using simple and clever series manipulations. It may be found at pp.$487-490$ (Sect. $6.48$).

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    $\begingroup$ The problem is stated in Sec. 4.48, with a hint (solution) in 5.48 (6.48). $\endgroup$
    – J.G.
    Commented Nov 9, 2022 at 19:18
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    $\begingroup$ @J.G. Yeap, since the problem statement is already given by OP (including the closed form), I guess he's only interested to jump right to the place with the solution. :-) $\endgroup$ Commented Nov 9, 2022 at 19:22

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