# An Euler sum with tails

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.

• If it was proposed $5$ years ago, please give the source! Commented Nov 9, 2022 at 19:06
• 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). Commented Nov 9, 2022 at 19:20
• The problem was proposed by Cornel Valean ... Commented Nov 10, 2022 at 7:57
• @Tolaso I agree with you! Thank you for the bounty! :-) Commented Nov 10, 2022 at 8:06

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. $$6.48$$).