# How to calculate the closed form of the series

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

• What do you mean exactly with $[x]$? Oct 27, 2014 at 9:25

$1)$ 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

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

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

• (+1) All the $\mathrm{Li}_2$'s and $\mathrm{Li}_3$'s leave me a bit unsettled, but perhaps these cannot be simplified away due to the odd nature of the question.
– robjohn
Oct 27, 2014 at 15:43
• $$\psi^{(1)}\left(\frac{2}{3}\right)+\psi^{(1)}\left(\frac{5}{6}\right)-\psi^{(1)}\left(\frac{1}{3}\right)-\psi^{(1)}\left(\frac{1}{6}\right) = -36\sqrt3\operatorname{Cl}_2\left(\frac{2\pi}{3}\right),$$ where $\operatorname{Cl}_2$ is Clausen's integral. For more details see my comment here and my question here. By the way the term "not the floor function" used for what? Which function then? Round? Oct 28, 2014 at 12:52
• @user153012 Thank you for the comment. I just edited my answer. Oct 28, 2014 at 13:04
• @user153012 Yeah, agree, I noticed that, but I preferred to let things like that. Oct 28, 2014 at 13:55
• May I politely ask if you could include some steps? I think these sort of answers might leave Cleo without a job (of course I am just kidding :P) +1 Oct 29, 2014 at 1:53