# 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.

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 $$\sum_{n\geq1}\frac{H_{n}^{\left(2\right)}}{n2^{n}}=3\textrm{Li}_{3}\left(\frac{1}{2}\right)-3\log\left(\frac{1}{2}\right)\textrm{Li}_{2}\left(\frac{1}{2}\right)-\log^{2}\left(\frac{1}{2}\right)\log\left(\frac{1}{2}\right)-2\zeta\left(3\right)=$$ $$=\frac{5\zeta\left(3\right)}{8}.$$

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 : $$S_m(z) = S_{2,m-1}(1)-S_{m-1,2}(1)-\left(S_{1,m-1}(1)-Li_m(1)\right) \log(z) - Li_m(1) \log(1-z) + \frac{(-1)^{m-1}}{(m-1)!} \int\limits_0^1 [\log(1-\xi)]^{m-1} \frac{\log[1-(1-\xi) z]}{\xi} d\xi$$ 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: $$S_m(z) = Li_{m+1}(z) + \sum\limits_{p=1}^{m_1} Li_p(z) Li_{m+1-p}(z) (-1)^{p-1} + (-1)^{m_1} \frac{1}{2} Li_{m_1+1}(z)^2$$ 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: $$r_{m_1}(z) := (-1)^{m_1} \int\limits_0^z \frac{Li_{m_1}(t)^2}{t} dt$$ 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}

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: $$\operatorname{Li}_3\left(\frac12\right)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$$

$$\operatorname{Li}_2\left(\frac12\right)=\frac12\zeta(2)-\frac12\ln^22$$

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 $$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n2^n}=\frac58\zeta(3)$$

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.