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We know that $$\sum _{k=1}^{\infty } \frac{H_k}{k^3} = \frac{\pi^4}{72}.$$ Is there a closed form for the sum $$S=\sum _{k=1}^{\infty } \frac{(-1)^k H_k}{k^3}?$$ Mathematica doesn't give anything resembling a closed form and I have no idea if one exists.

user111187
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    $$\displaystyle \sum_{k=1}^{\infty} \frac{H_k}{k^3} = \sum_{n,k=1}^{\infty}\frac{1}{nk^2(n+k)} = \frac{1}{2}\left(\sum_{n,k=1}^{\infty}\frac{1}{nk^2(n+k)}+ \sum_{n,k=1}^{\infty}\frac{1}{n^2k(n+k)}\right) \= \frac{1}{2}\sum_{n,k=1}^{\infty}\frac{1}{n^2k^2} = \frac{1}{2}\left(\sum_{n=1}^{\infty}\frac{1}{n^2}\right)^2 = \frac{\pi^4}{72}$$ and for the second part see here. – r9m Jan 05 '15 at 10:19
  • See my answer and follow the links and you will find useful techniques. – Mhenni Benghorbal Jan 05 '15 at 10:23
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    Generally speaking, \begin{align} \sum^\infty_{n=1}\frac{H_n}{n^3}z^n =&2{\rm Li}_4(z)+{\rm Li}_4\left(\frac{z}{z-1}\right)-{\rm Li}_4(1-z)-{\rm Li}_3(z)\ln(1-z)-\frac{1}{2}{\rm Li}_2^2\left(\frac{z}{z-1}\right)\ &+\frac{1}{2}{\rm Li}_2(z)\ln^2(1-z)+\frac{1}{2}{\rm Li}_2^2(z)+\frac{1}{6}\ln^4(1-z)-\frac{1}{6}\ln{z}\ln^3(1-z)\ &+\frac{\pi^2}{12}\ln^2(1-z)+\zeta(3)\ln(1-z)+\frac{\pi^4}{90} \end{align} – M.N.C.E. Jan 05 '15 at 11:56
  • @M.N.C.E. Thanks, it must have been a pain to TEX that. – user111187 Jan 05 '15 at 12:02
  • @M.N.C.E. not safe to have $\frac{z}{z-1}$ when lhs series converges for $z = 1$ but rhs does not make sense, best remove euler-transforms and keep $$\displaystyle \sum\limits_{n=1}^{\infty} \dfrac{H_n}{n^3}x^n = -\frac{1}{2}\sum\limits_{n=1}^{\infty}\frac{1}{n^2}\sum\limits_{k=1}^{n}\frac{(1-x)^k}{k^2} -\frac{1}{2} \zeta(2)\operatorname{Li}_2(x)+\frac{7}{8}\zeta(4)-\frac{1}{4}\operatorname{Li}_2^2(1-x)+\frac{1}{4}\zeta^2(2)+\operatorname{Li}_4(x)+\frac{1}{4}\log^2 x\log^2(1-x)+\frac{1}{2}\log x\log (1-x)\operatorname{Li}_2(1-x)+\zeta(3)\log x -\log x\operatorname{Li}_2(1-x)$$ – r9m Jan 10 '15 at 22:17

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Since: $$ \sum_{k\geq 1} H_k x^k = \frac{-\log(1-x)}{1-x} $$ we have: $$ \sum_{k\geq 1} \frac{H_k}{k+1} x^{k+1} = \frac{1}{2}\log^2(1-x), $$

$$ \sum_{k\geq 1} \frac{(-1)^{k+1}H_k}{k+1} x^{k} = \frac{1}{2x}\log^2(1+x), $$ and since $\int_{0}^{1} x^k\log(x) = -\frac{1}{(k+1)^2}$, it follows that: $$ S=\sum_{k\geq 1} \frac{(-1)^{k}H_k}{(k+1)^3}=\int_{0}^{1}\frac{\log^2(1+x)\log x}{2x}\,dx \tag{1}$$ for the last integral, Mathematica returns: $$ S = \frac{\pi ^4}{48}-\frac{\log^4 2}{12}+\frac{\pi^2\log^2 2}{12} -2 \operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{7}{4}\zeta(3)\log 2 \tag{2}$$ and since: $$ \sum_{k\geq 1}\frac{(-1)^{k}\frac{1}{k+1}}{(k+1)^3}=-1+\frac{7\pi^4}{720}\tag{3} $$ it follows that:

$$ \sum_{k\geq 1}\frac{(-1)^k H_k}{k^3}=-\frac{11\pi ^4}{360}+\frac{\log^4 2}{12}-\frac{\pi^2\log^2 2}{12} +2 \operatorname{Li}_4\left(\frac{1}{2}\right)+\frac{7}{4}\zeta(3)\log 2.\tag{4}$$

Jack D'Aurizio
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