Strategies for evaluating sums $\sum_{n=1}^\infty \frac{H_n^{(m)}z^n}{n}$ I'm looking for strategies for evaluating the following sums for given $z$ and $m$:
$$
\mathcal{S}_m(z):=\sum_{n=1}^\infty \frac{H_n^{(m)}z^n}{n},
$$
where $H_n^{(m)}$ is the generalized harmonic number, and $|z|<1$, $m \in \mathbb{R}$.
Using the generating function of the generalized harmonic numbers, an equivalent problem is to evaluate the following integral:
$$
\mathcal{S}_m(z) = \int_0^z \frac{\operatorname{Li}_m(t)}{t(1-t)}\,dt,
$$
where $\operatorname{Li}_m(t)$ is the polylogarithm function, and $|z|<1$, $m \in \mathbb{R}$.

Question 1: Are there any way to reduce the sum to Euler sum values, given by Flajolet–Salvy paper?
Question 2: Are there any way to reduce the integral to integrals given by Freitas paper?


The case $m=1$ and $z=1/2$ was the problem 1240 in Mathematics Magazine, Vol. 60, No. 2, pp. 118–119. (Apr., 1987) by Coffman, S. W.
$$
\mathcal{S}_1\left(\tfrac12\right)=\sum_{n=1}^\infty \frac{H_n}{n2^n} = \frac{\pi^2}{12}.
$$
There are several solutions in the linked paper.
The more interesting case $m=2$ and $z=1/2$ is listed at Harmonic Number, MathWorld, eq. $(41)$:
$$
\mathcal{S}_2\left(\tfrac12\right)=\sum_{n=1}^\infty \frac{H_n^{(2)}}{n2^n} = \frac{5}{8}\zeta(3).
$$
We know less about the evaluation. At the MathWorld it is marked as "B. Cloitre (pers. comm., Oct. 4, 2004)". This value is also listed at pi314.net, eq. $(701)$. Unfortunately, I don't know about any paper/book reference for this value. It would be nice to see some.

Question 3: How could we evaluate the case $m=2$, $z=1/2$?

It would be nice to see a solution for the sum form, but also solutions for the integral form are welcome.
 A: By using the integral representation of the polylogarithm , inserting it into the integral in question, changing the order of integration , integrating over $t$ and then using Abel's identity for the dilogarithm we obtained the following :
\begin{equation}
S_m(z) = S_{2,m-1}(1)-S_{m-1,2}(1)-\left(S_{1,m-1}(1)-Li_m(1)\right) \log(z) - Li_m(1) \log(1-z) + \frac{(-1)^{m-1}}{(m-1)!} \int\limits_0^1 [\log(1-\xi)]^{m-1} \frac{\log[1-(1-\xi) z]}{\xi} d\xi
\end{equation}
where $S_{n,m}(1)$ are the Nielsen generalized polylogarithms at unity. I guess that integration by parts will be now the best strategy to evaluate the integral on the right hand side.
On the other hand assume that $m=2 m_1+1$ is odd. Then by decomposing the denominator into simple fractions and then integrating by parts we have:
\begin{eqnarray}
S_m(z) 
&=& Li_{m+1}(z) + \int\limits_0^z \frac{Li_m(t)}{1-t} dt \\
&=& Li_{m+1}(z) + Li_1(z) Li_m(z) - \int\limits_0^z Li_1(t) \frac{Li_{m-1}(t)}{t} dt \\
&\vdots&\\
&=& Li_{m+1}(z) + \sum\limits_{p=1}^{P} Li_p(z) Li_{m+1-p}(z) (-1)^{p-1} + (-1)^P \int\limits_0^z Li_P(t) \frac{Li_{m-P}(t)}{t} dt 
\end{eqnarray}
Now we choose $p=m_1$ and in the last integral the integrand is clearly a full derivative. Therefore we have:
\begin{equation}
S_m(z) = Li_{m+1}(z) + \sum\limits_{p=1}^{m_1} Li_p(z) Li_{m+1-p}(z) (-1)^{p-1} + (-1)^{m_1} \frac{1}{2} Li_{m_1+1}(z)^2
\end{equation}
Unfortunately when $m$ is even this method doesn't work. How do we handle the even case then ?
Now let us assume that $m=2 m_1$ is even. Here the residual integral reads:
\begin{equation}
r_{m_1}(z) := (-1)^{m_1} \int\limits_0^z \frac{Li_{m_1}(t)^2}{t} dt 
\end{equation}
Firstly we handle the case $m_1=1$. Here we just integrate by parts. We have:
\begin{eqnarray}
-r_1(z) &=& \int\limits_0^z \frac{[\log(1-t)]^2}{t} dt\\
&=& \log(z) [\log(1-z)]^2 + \int\limits_0^z \log(t)\cdot \frac{2 \log(1-t)}{1-t} dt\\
&=&  \log(z) [\log(1-z)]^2 + Li_2(1-z)\cdot 2 \log(1-z) + \int\limits_0^z Li_2(1-t) \cdot \frac{2}{1-t} dt \\
&=&  \log(z) [\log(1-z)]^2 + Li_2(1-z)\cdot 2 \log(1-z) +  2\left[Li_3(1) - Li_3(1-z)\right]
\end{eqnarray}
A: Using generating function :
$$\sum_{n=1}^\infty\frac{x^nH_n^{(2)}}{n}=\operatorname{Li}_3(x)+2\operatorname{Li}_3(1-x)-\ln(1-x)\operatorname{Li}_2(1-x)-\zeta(2)\ln(1-x)-2\zeta(3)$$
and by taking $x=1/2$, we have 
$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n2^n}=3\operatorname{Li}_3\left(\frac12\right)+\ln2\operatorname{Li}_2\left(\frac12\right)+\ln2\zeta(2)-2\zeta(3)$$
Substituting :
$$\operatorname{Li}_3\left(\frac12\right)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$$
$$\operatorname{Li}_2\left(\frac12\right)=\frac12\zeta(2)-\frac12\ln^22$$
We get $$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n2^n}=\frac58\zeta(3)$$
A: Different approach:
Using the fact that $$\sum_{n=1}^\infty H_n^{(2)}x^n=\frac{\operatorname{Li}_2(x)}{1-x}$$
divide both sides by $x$ then integrate from $x=0$ to $1/2$, we get
\begin{align}
S&=\sum_{n=1}^\infty\frac{H_n^{(2)}}{n2^n}=\int_0^{1/2}\frac{\operatorname{Li}_2(x)}{x(1-x)}\ dx\\
&=\int_0^{1/2}\frac{\operatorname{Li}_2(x)}{x}\ dx+\int_0^{1/2}\frac{\operatorname{Li}_2(x)}{1-x}\ dx\\
&=\operatorname{Li}_3\left(\frac12\right)-\left.\ln(1-x)\operatorname{Li}_2(x)\right|_0^{1/2}-\int_0^{1/2}\frac{\ln^2(1-x)}{x}\ dx\\
&=\operatorname{Li}_3\left(\frac12\right)+\ln2\operatorname{Li}_2\left(\frac12\right)-\int_{1/2}^{1}\frac{\ln^2x}{1-x}\ dx\\
&=\operatorname{Li}_3\left(\frac12\right)+\ln2\operatorname{Li}_2\left(\frac12\right)-\sum_{n=1}^\infty\int_{1/2}^1 x^{n-1}\ln^2x\ dx\\
&=\operatorname{Li}_3\left(\frac12\right)+\ln2\operatorname{Li}_2\left(\frac12\right)-\sum_{n=1}^\infty\left(\frac{2}{n^3}-\frac{\ln^22}{n2^n}-\frac{2\ln2}{n^2 2^n}-\frac{2}{n^3 2^n}\right)\\
&=\operatorname{Li}_3\left(\frac12\right)+\ln2\operatorname{Li}_2\left(\frac12\right)-2\zeta(3)+\ln^32+2\ln2\operatorname{Li}_2\left(\frac12\right)+2\operatorname{Li}_3\left(\frac12\right)\\
&=3\operatorname{Li}_3\left(\frac12\right)+3\ln2\operatorname{Li}_2\left(\frac12\right)-2\zeta(3)+\ln^32\\
&=\frac58\zeta(3)
\end{align}
where we used in our calculations:
$$\operatorname{Li}_3\left(\frac12\right)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$$
$$\operatorname{Li}_2\left(\frac12\right)=\frac12\zeta(2)-\frac12\ln^22$$
A: About quesiton 3, we have $$\sum_{n\geq1}\frac{H_{n}^{\left(2\right)}}{n2^{n}}=\int_{0}^{1/2}\frac{\textrm{Li}_{2}\left(x\right)}{x\left(1-x\right)}dx=\int_{0}^{1/2}\frac{\textrm{Li}_{2}\left(x\right)}{x}dx+$$ $$+\int_{0}^{1/2}\frac{\textrm{Li}_{2}\left(x\right)}{1-x}dx=\textrm{Li}_{3}\left(\frac{1}{2}\right)+\int_{0}^{1/2}\frac{\textrm{Li}_{2}\left(x\right)}{1-x}dx
 $$ now note that, using integration by parts $$\int_{0}^{1/2}\frac{\textrm{Li}_{2}\left(x\right)}{1-x}dx=-\log\left(\frac{1}{2}\right)\textrm{Li}_{2}\left(\frac{1}{2}\right)-\int_{0}^{1/2}\frac{\log^{2}\left(1-x\right)}{x}dx.$$
 So let analyze $$J=\int_{0}^{1/2}\frac{\log^{2}\left(1-x\right)}{x}dx
 $$ using integration by parts few times $$J=\log^{2}\left(\frac{1}{2}\right)\log\left(\frac{1}{2}\right)+2\int_{0}^{1/2}\frac{\log\left(1-x\right)\log\left(x\right)}{1-x}dx=
 $$ $$=\log^{2}\left(\frac{1}{2}\right)\log\left(\frac{1}{2}\right)+2\textrm{Li}_{2}\left(\frac{1}{2}\right)\log\left(\frac{1}{2}\right)+2\int_{1/2}^{1}\frac{\textrm{Li}_{2}\left(u\right)}{u}du
 $$ $$=\log^{2}\left(\frac{1}{2}\right)\log\left(\frac{1}{2}\right)+2\textrm{Li}_{2}\left(\frac{1}{2}\right)\log\left(\frac{1}{2}\right)+2\zeta\left(3\right)-2\textrm{Li}_{3}\left(\frac{1}{2}\right)
 $$ so we have $$\sum_{n\geq1}\frac{H_{n}^{\left(2\right)}}{n2^{n}}=3\textrm{Li}_{3}\left(\frac{1}{2}\right)-3\log\left(\frac{1}{2}\right)\textrm{Li}_{2}\left(\frac{1}{2}\right)-\log^{2}\left(\frac{1}{2}\right)\log\left(\frac{1}{2}\right)-2\zeta\left(3\right)=
 $$ $$=\frac{5\zeta\left(3\right)}{8}.
 $$
A: By Cauchy product we have 
$$-\ln(1-x)\operatorname{Li}_2(x)=\sum_{n1}^\infty\left(\frac{2H_n}{n^2}+\frac{H_n^{(2)}}{n}-\frac3{n^3}\right)x^n$$
and by setting $x=1/2$ and using $\sum_{n=1}^\infty\frac{H_n}{n^22^n}=\zeta(3)-\frac12\ln2\zeta(2)$ we get the desired closed form.
