TITLE: Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^5 2^n}$
QUESTION [10 upvotes]: 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.

REPLY [5 votes]: $\newcommand{\Li}{\mathrm{Li}}$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.