How to calculate the closed form of the series

Question

We know that the closed form of the series $$\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k=-\frac{11\pi^4}{360}+\frac{\ln^42-\pi^2\ln^22}{12}+2\mathrm{Li}_4\left(\frac12\right)+\frac{7\ln 2}{4}\zeta(3),$$ but how to evaluate the following series $$\sum\limits_{n = 1}^\infty {\frac{{{H_{\left[ {\frac{n}{3}} \right]}}}}{{{n^2}}}{{\left( { - 1} \right)}^{n - 1}}} ,\sum\limits_{n = 1}^\infty {\frac{{{H_{\left[ {\frac{n}{3}} \right]}}}}{{{n^3}}}{{\left( { - 1} \right)}^{n - 1}}} .$$

Answer 1

$1)$ The case where $\left[ x \right]$ is not considered floor function $\left(\left[ x \right]=x \right)$

$$\sum\limits_{n = 1}^\infty {\frac{{{H_{\left[ {\frac{n}{3}} \right]}}}}{{{n^2}}}{{\left( { - 1} \right)}^{n - 1}}} =$$ $$\frac{1}{12}\log^3(3)-\frac{\pi^2}{36}\log\left(\frac{256}{243}\right)-\frac{7}{24}\zeta(3)-\frac{1}{72}\ln(3)\left(9\ln^2(3)-5\pi^2\right)$$ $$+\operatorname{Li_3}\left(\frac{1}{6}\left(3+i\sqrt{3}\right)\right)+\operatorname{Li_3}\left(\frac{1}{6}\left(3-i\sqrt{3}\right)\right)+$$

$$i\frac{\pi}{6}\left(\operatorname{Li_2}\left(\frac{1}{6}\left(3-i\sqrt{3}\right)\right)-\operatorname{Li_2}\left(\frac{1}{6}\left(3+i\sqrt{3}\right)\right)\right)+i\frac{\pi}{3}\left(\operatorname{Li_2}\left(\frac{1}{4}\left(3+i\sqrt{3}\right)\right)-\operatorname{Li_2}\left(\frac{1}{4}\left(3-i\sqrt{3}\right)\right)\right)$$

$2)$ The case where $\left[ x \right]$ is considered floor function

$$\sum\limits_{n = 1}^\infty {\frac{{{H_{\left[ {\frac{n}{3}} \right]}}}}{{{n^2}}}{{\left( { - 1} \right)}^{n - 1}}} =$$

$$\frac{161}{72}\zeta(3)+\frac{\pi^2}{27}\log(3)+\frac{\pi}{72\sqrt{3}}\left(\underbrace{\psi^{(1)}\left(\frac{2}{3}\right)+\psi^{(1)}\left(\frac{5}{6}\right)-\psi^{(1)}\left(\frac{1}{3}\right)-\psi^{(1)}\left(\frac{1}{6}\right)}_{\displaystyle -36\sqrt3\operatorname{Cl}_2\left(\frac{2\pi}{3}\right)}\right)$$ $$+i\frac{5\pi}{9}\operatorname{Li_2}\left(1-i\frac{\sqrt{3}}{3}\right)-i\frac{5\pi}{9}\operatorname{Li_2}\left(1+i\frac{\sqrt{3}}{3}\right)$$ $$+i\frac{7\pi}{9}\operatorname{Li_2}\left(\frac{1}{6}\left(3+i\sqrt{3}\right)\right)-i\frac{7\pi}{9}\operatorname{Li_2}\left(\frac{1}{6}\left(3-i\sqrt{3}\right)\right)$$