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where $H_n$ is the harmonic number and can be defined as:

$H_n=1+\frac12+\frac13+...+\frac1n$

$H_n^{(3)}=1+\frac1{2^3}+\frac1{3^3}+...+\frac1{n^3}$

I managed to prove $\quad\displaystyle\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}=\frac{21}{32}\zeta(5)-\frac34\zeta(2)\zeta(3)\quad$ using logarithmic integral and couple harmonic identities. other approaches are much appreciated.

you can find a different solution by Cornel in his book (Almost impossible integrals, sum and series).

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  • $\begingroup$ @ Ali Shather Over the last couple of weeks you have provided interesting results for special cases. What about explaining your general method and showing its capabiities in solving a class of problems, e.g. for $S(p,q) = \sum_{k=1}^\infty (-1)^k H_{k}^{(p)}/k^q$ and answer questins such as "Which combinations of $,p$ and $q$ permit closed expressions?" and "What is new in these results?" $\endgroup$ – Dr. Wolfgang Hintze Jun 8 at 16:40
  • $\begingroup$ @ Dr. Wolfgang Hintze thank you. actually I solved these sums over a year ago and I posted some of them on my facebook group ( harmonic series) but I've been posting them here lately as I've just learned LaTex. about your question, I know that Bastien derived the generalization for S(1,q) here arxiv.org/pdf/… where q is even. $\endgroup$ – Ali Shather Jun 8 at 18:04
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    $\begingroup$ @ Dr. Wolfgang and couple days ago Cornel derived the generalization for S( p,1) here researchgate.net/publication/…. The generalization of S(p,q) is extremely hard and its not in the mathematical literature yet. I do not see myself good enough to investigate it and I am just focused on providing new solutions to particular cases. $\endgroup$ – Ali Shather Jun 8 at 18:04
  • $\begingroup$ @ Ali Shather From the work of Bastien (thanks for the hint) I conclude that $S(1,5)=\sum_{k=1}^{\infty} (-1)^k \frac{H_{k}}{k^5}$ is the first sum which seems not to have a closed form (in terms). So this is a real challenge. $\endgroup$ – Dr. Wolfgang Hintze Jun 8 at 21:56
  • $\begingroup$ @Dr.WolfgangHintze no problem. Yes it's pretty tough and that's why mathematicians still struggle with it. $\endgroup$ – Ali Shather Jun 8 at 22:14
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using the fact that $\displaystyle\sum_{n=1}^\infty z^nH_n^{(3)}=\frac{\operatorname{Li}_3(z)}{1-z}\ $ divide both sides by $z$ then integrate from $z=0$ to $x$, we get $$\sum_{n=1}^\infty \frac{x^nH_n^{(3)}}{n}=\operatorname{Li}_4(x)-\ln(1-x)\operatorname{Li}_3(x)-\frac12\operatorname{Li}_2^2(x)\tag{1}$$

replace $x$ with $-x$ in(1), then divide both sides by $x$ and integrate from $0$ to $1$, we get

\begin{align} \sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}&=\operatorname{Li}_5(-1)-\underbrace{\int_0^1\frac{\ln(1+x)\operatorname{Li}_3(-x)}{x}\ dx}_{IBP}-\frac12\int_0^1\frac{\operatorname{Li}_2^2(-x)}{x}\ dx\\ &=\operatorname{Li}_5(-1)+\operatorname{Li}_2(-1)\operatorname{Li}_3(-1)-\frac32\int_0^1\frac{\operatorname{Li}_2^2(-x)}{x}\ dx\\ &=\frac38\zeta(2)\zeta(3)-\frac{15}{16}\zeta(5)-\frac32\int_0^1\frac{\operatorname{Li}_2^2(-x)}{x}\ dx\\ \end{align} I proved here \begin{align} \int_0^1\frac{\operatorname{Li}_2^2(-x)}{x}\ dx=\frac34\zeta(2)\zeta(3)-\frac{17}{16}\zeta(5) \end{align} which follows that

$$\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}=\frac{21}{32}\zeta(5)-\frac34\zeta(2)\zeta(3)$$


BONUS:

By setting $x=-1$ in (1) we have

$$\sum_{n=1}^\infty(-1)^n\frac{H_n^{(3)}}{n}=\frac34\ln2\zeta(3)-\frac{19}{16}\zeta(4)$$

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