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

xuce1234
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1 Answers1

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$1)$ 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

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

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

user 1591719
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  • (+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 '14 at 15:43
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    $$\psi^{(1)}\left(\frac{2}{3}\right)+\psi^{(1)}\left(\frac{5}{6}\right)-\psi^{(1)}\left(\frac{1}{3}\right)-\psi^{(1)}\left(\frac{1}{6}\right) = -36\sqrt3\operatorname{Cl}_2\left(\frac{2\pi}{3}\right),$$ 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? – user153012 Oct 28 '14 at 12:52
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    @user153012 Thank you for the comment. I just edited my answer. – user 1591719 Oct 28 '14 at 13:04
  • @user153012 Yeah, agree, I noticed that, but I preferred to let things like that. – user 1591719 Oct 28 '14 at 13:55
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    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 – M.N.C.E. Oct 29 '14 at 01:53