# Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^5 2^n}$

Given the nth harmonic number $$H_n = \sum_{j=1}^{n} \frac{1}{j}$$, we get from this post that apparently,

$$\sum_{n=1}^{\infty}\frac{H_n}{n^k}z^n= S_{k-1,2}(z) + \rm{Li}_{\,k+1}(z)$$

for $$-1\leq z\leq 1$$, and with Nielsen generalized polylogarithm $$S_{n,p}(z)$$ and polylogarithm $$\rm{Li}_n(z)$$. Hence for small $$k$$,

$$\sum_{n=1}^{\infty}\frac{H_n}{n^2\, 2^n}= S_{1,2}\big(\tfrac12\big)+\rm{Li}_3\big(\tfrac12\big)$$

$$\sum_{n=1}^{\infty}\frac{H_n}{n^3\, 2^n}= S_{2,2}\big(\tfrac12\big)+\rm{Li}_4\big(\tfrac12\big)$$

$$\sum_{n=1}^{\infty}\frac{H_n}{n^4\, 2^n}= S_{3,2}\big(\tfrac12\big)+\rm{Li}_5\big(\tfrac12\big)$$

and so on. Explicitly, given $$a=\ln 2$$,

$$S_{1,2}\big(\tfrac12\big) +\tfrac1{6}a^3-\tfrac18 \zeta(3)=0$$

$$S_{2,2}\big(\tfrac12\big) +\tfrac1{168}a^4+\tfrac17a^2\,\rm{Li}_2\big(\tfrac12\big)+\tfrac17a\,\rm{Li}_3\big(\tfrac12\big)-\tfrac18\zeta(4) = 0$$

which are discussed in this and this post. And by yours truly,

$$S_{3,2}\big(\tfrac12\big) -A+B = 0$$

$$A = \tfrac{41}{840}a^5+\tfrac5{21}a^3\,\rm{Li}_2\big(\tfrac12\big)+\tfrac47a^2\,\rm{Li}_3\big(\tfrac12\big)+a\,\rm{Li}_4\big(\tfrac12\big) + \rm{Li}_5\big(\tfrac12\big)$$

$$B=\tfrac12\zeta(2)\zeta(3)+\tfrac18a\,\zeta(4)-\tfrac1{32}\zeta(5)$$

Q: What, however, is the explicit evaluation in ordinary polylogs of the next steps, namely $$S_{4,2}\big(\tfrac12\big)$$ and $$S_{5,2}\big(\tfrac12\big)$$?

P.S. Try as I might, they resist being evaluated and there are indications these higher order integrals may not be expressible by ordinary polylogs.

• Also relevant: math.stackexchange.com/questions/944065/… – nospoon Jun 1 '19 at 9:22
• @nospoon: Ah, that's the case $k=4$. Didn't see that. – Tito Piezas III Jun 1 '19 at 9:24
• @nospoon: If there is a recursion involved (as the answer in your link seems to suggest), then it may be possible to find all subsequent steps. – Tito Piezas III Jun 1 '19 at 9:30
• Another relevant link for $k=5$. math.stackexchange.com/questions/972775/… – nospoon Jun 2 '19 at 20:57
• @nospoon: Thanks for the link. At first, I thought a closed-form for $k=5$ was found, but it just expresses it in several Nielsen polylogarithms. Kindly see this new post. – Tito Piezas III Jun 3 '19 at 8:07

## 2 Answers

$$\newcommand{\Li}{\mathrm{Li}}$$You are probably correct that the sufficiently higher order sums do not have a closed form in terms of classical polylogarithms. In particular the $$k=6$$ sum should not have such an expression, although this claim rests on some big conjectures (the so-called Grothendeick Period Conjecture, or some sufficiently special-case). However the $$k=5$$ case does appear to have a closed form, as given below.

Essentially there is an algebraically defined version of (Nielsen) polylogs, the so-called motivic (Nielsen) polylogs. These objects have a much richer structure, and there is an invariant (the coproduct $$\Delta$$ and the cobracket $$\delta$$) which can distinguish Nielsen from classical polylogs. The cobracket $$\delta\Li_n(x) = 0$$, whereas $$\delta S_{5,2}(x) = \Li_2(x) \wedge \Li_5(1) + \Li_4(x) \wedge \Li_3(1) \neq 0$$; the factors of the wedge are viewed modulo products. In particular, $$\delta S_{5,2}(\frac{1}{2}) = \Li_4(\frac{1}{2}) \wedge \zeta(3) \neq 0$$. (Can check that $$L_4(\frac{1}{2}) / \zeta(4) \not\in \mathbb{Q}$$, where $$L_4$$ is some single-valued version of $$\Li_4$$.)

However $$\delta S_{4,2}(x) = \Li_3(x) \wedge \Li_3(1)$$, and since $$\Li_3(\frac{1}{2}) = \frac{7}{8} \zeta(3) - \frac{1}{2} \zeta(2) \log(2) + \frac{1}{6} \log(2)^3$$, one gets $$\delta S_{4,2}\Big(\frac{1}{2}\Big) = \frac{7}{8} \zeta(3) \wedge \zeta(3) = 0 \,,$$ by the antisymmetry of the wedge. Another Big Conjecture from Goncharov claims that classical polylogs are exactly the kernel of this cobracket, so conjecturally we should be able to expression $$S_{4,2}(\frac{1}{2})$$ in terms of $$\Li_6$$.

In the answer I posted to another of your questions, I gave an explicit formula for $$S_{4,2}(-1)$$ and some explanation via the coproduct of how one obtains it (up to a rational numerically determined coefficient of $$\zeta(6)$$). One can also get a formula for $$S_{4,2}(\frac{1}{2})$$, and hence for sum for $$k=5$$, from the $$S_{4,2}(-1)$$ reduction. In particular \begin{align*} \sum_{n=1}^\infty\frac{(-1)^n H_n}{n^5} = {} & S_{4,2}(-1) - \frac{31}{32} \zeta(6) = \zeta(\overline{5}, 1) - \frac{31}{32} \zeta(6) \\ \overset{?}{=} {} & \frac{1}{13} \bigg(\frac{1}{3}\Li_6\Big(-\frac{1}{8}\Big)-162 \Li_6\Big(-\frac{1}{2}\Big)-126 \Li_6\Big(\frac{1}{2}\Big)\bigg) -\frac{4783 }{1248}\zeta (6) +\frac{3}{8} \zeta (3)^2 \\ & {}+\frac{31}{16} \zeta (5) \log(2) -\frac{15}{26} \zeta (4) \log ^2(2) +\frac{3}{104} \zeta (2) \log^4(2) -\frac{1}{208} \log ^6(2) \,, \\[2em] \sum_{n=1}^\infty\frac{H_n}{n^5 2^n} \overset{?}{=} {} & -\frac{1}{26} \left( \frac{1}{3}\Li_6\left(-\frac{1}{8}\right) -162\Li_6\left(-\frac{1}{2}\right) -204\Li_6\left(\frac{1}{2}\right)\right) -\frac{101 }{624} \zeta (6) -\frac{7}{16} \zeta (3)^2 \\ &+\Big(\Li_5\left(\frac{1}{2}\right) -\zeta (5)+\frac{1}{2}\zeta (2) \zeta (3) \Big) \log (2) +\frac{73}{208} \zeta (4) \log ^2(2)\\ &-\frac{1}{6} \zeta (3) \log^3(2) +\frac{17}{624} \zeta (2) \log ^4(2) -\frac{11}{6240}\log ^6(2) \,. \end{align*} Amusingly, since $$\Li_3(\phi^{-2}) = \frac{4}{5} \zeta(3) + \text{products}$$, where $$\phi = \frac{1 + \sqrt{5}}{2}$$ is the golden ratio, we can also conjecturally evaluate $$S_{4,2}(\phi^{-2})$$ (up to a rational but numerically determined coefficient of $$\zeta(6)$$). We have \begin{align*} \sum_{n=1}^\infty\frac{H_n}{n^5 \phi^{2n}} \overset{?}{=} {} & \frac{1}{396} \Big( 2 \Li_6\big( \phi ^{-6} \big) -128 \Li_6\big( \phi ^{-3}\big) +1197\Li_6\big(\phi ^{-2}\big) -576 \Li_6\big(\phi^{-1} \big) \Big) \\ &+\frac{35 }{99}\zeta (6) +\frac{2}{5} \zeta (3)^2 + \Li_5 \left(\phi ^{-2}\right) \log (\phi) -\zeta (5)\log (\phi ) \\ & +\frac{2}{11} \zeta (4) \log ^2(\phi ) -\zeta (3) \Li_3\left(\phi ^{-2}\right) +\frac{10}{33} \zeta (2) \log ^4(\phi ) -\frac{79}{990} \log ^6(\phi ) \,. \end{align*} The only uncertainty in these formulae should be the (necessarily rational) coefficient of $$\zeta(6)$$, all other terms are fixed by analysis of the coproduct.

Let $$I$$ denotes $$\int_0^1\frac{\ln^4(1+x)\ln x}{x}\ dx$$

I proved here

\begin{align} I&=-120\operatorname{Li}_6\left(\frac12\right)-72\ln2\operatorname{Li}_5\left(\frac12\right)-24\ln^22\operatorname{Li}_4\left(\frac12\right)+78\zeta(6)+\frac34\ln2\zeta(5)\\ &\quad-\frac32\ln^22\zeta(4)-3\ln^32\zeta(3)+2\ln^42\zeta(2)+12\zeta^2(3)-12\ln2\zeta(2)\zeta(3)\\ &\quad-\frac{17}{30}\ln^62+24\sum_{n=1}^\infty\frac{H_n}{n^52^n}\tag{1} \end{align}

On the other hand and by applying integration by parts we have

\begin{align} I=-2\int_0^1\frac{\ln^3(1+x)\ln^2x}{1+x}\ dx \end{align} You can find here the following identity

$$-\frac{\ln^3(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$

Replacing $$x$$ with $$-x$$ yields

$$-\frac{\ln^3(1+x)}{1+x}=\sum_{n=1}^\infty (-1)^nx^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$

Now we can write \begin{align} I&=2\sum_{n=1}^\infty (-1)^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)\int_0^1 x^n \ln^2x\ dx\\ &=-2\sum_{n=1}^\infty (-1)^n\left(H_{n-1}^3-3H_{n-1}H_{n-1}^{(2)}+2H_{n-1}^{(3)}\right)\int_0^1 x^{n-1} \ln^2x\ dx\\ &=-2\sum_{n=1}^\infty (-1)^n\left(H_{n}^3-3H_{n}H_{n}^{(2)}+2H_{n}^{(3)}+\frac{6H_n}{n^3}-\frac{3H_n^2}{n}+\frac{3H_n^{(2)}}{n}-\frac{6}{n^2}\right)\left(\frac{2}{n^3}\right)\\ \end{align}

Then

\begin{align} I&=12\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^3}-4\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^3}-8\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^3}-24\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}\\ &\quad+12\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^4}-12\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^4}+24\sum_{n=1}^\infty\frac{(-1)^n}{n^6}\tag{2} \end{align}

We can see from $$(1)$$ and $$(2)$$ that our target sum is related to $$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}$$.

I am not sure if the alternating sums in $$(2)$$ have closed form or not but I am sure that $$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}$$ has no closed form because the power of the denominator is odd more than 3. So the target sum has no closed form unless $$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}$$ cancels out with other sums somehow which I doubt.