Cómo calcular la forma cerrada de la serie

Question

Sabemos que la forma cerrada de la serie $$\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k=-\frac{11\pi^4}{360}+\frac{\ln^42-\pi^2\ln^22}{12}+2\mathrm{Li}_4\left(\frac12\right)+\frac{7\ln 2}{4}\zeta(3),$$ pero cómo evaluar la siguiente serie $$\sum\limits_{n = 1}^\infty {\frac{{{H_{\left[ {\frac{n}{3}} \right]}}}}{{{n^2}}}{{\left( { - 1} \right)}^{n - 1}}} ,\sum\limits_{n = 1}^\infty {\frac{{{H_{\left[ {\frac{n}{3}} \right]}}}}{{{n^3}}}{{\left( { - 1} \right)}^{n - 1}}} .$$

Answer 1

$1)$ El caso en el que $\left[ x \right]$ no es función del piso considerado $\left(\left[ x \right]=x \right)$

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

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

$2)$ El caso en el que $\left[ x \right]$ es función del piso considerado

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

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