So I have been working with some power series involving the Harmonic numbers. I have been able to evaluate the first couple of sums as
$$f(z,0)=\sum_{n=1}^\infty H_n z^n=\sum_{n=1}^\infty z^n\int_0^1 \frac{1-x^n}{1-x}dx=-\frac{\ln(1-z)}{1-z}$$
$$f(z,1)=\sum_{n=1}^\infty \frac{H_n}{n}z^n=\int_0^z\frac{f(x,0)}{x}dx=\operatorname{Li}_2(z)+\frac{\ln^2(1-z)}{2}$$
$$f(z,2)=\sum_{n=1}^\infty \frac{H_n}{n^2}z^n=\int_0^z \frac{f(x,1)}{x}dx=\operatorname{Li}_3(z)-\operatorname{Li}_3(1-z)+\operatorname{Li}_2(1-z)\ln(1-z)+\frac{\ln(z)\ln^2(1-z)}{2}+\zeta(3)$$
and in general$$f(z,a)=\sum_{n=1}^\infty \frac{H_n}{n^a}z^n=\int_0^z \frac{f(x,a-1)}{x}dx$$ I was wondering if it is possible to derive a general formula for $f(z,a)$ as computing each integral of $\frac{f(x,a-1)}{x}$ becomes very tedious and complicated to do and does not seem to provide a closed form answer for any $a$. I know that there exists a closed form for $f(1,a)$ but how could we generalize this to $f(z,a)$?
Closed form for $f(z,a)=\sum_{n=1}^\infty \frac{H_n}{n^a}z^n$
definite integralsharmonic-numberspolylogarithmsequences-and-series
Related Solutions
To calculate these two sums, we are going to establish two relations and solve them by elimination.
To establish the first relation, we use $\displaystyle I=\int_0^1\frac{\ln^4(1+x)+6\ln^2(1-x)\ln^2(1+x)}{x}\ dx=\frac{21}4\zeta(5)\tag{1}$
which was proved by Khalef Ruhemi ( unfortunately he is not an MSE user).
The proof as follows: using the algebraic identity $\ b^4+6a^2b^2=\frac12(a-b)^4+\frac12(a+b)^4-a^4$
with $\ a=\ln(1-x)$ and $\ b=\ln(1+x)$ , divide both sides by $x$ then integrate, we get
$$I=\frac12\underbrace{\int_0^1\frac1x{\ln^4\left(\frac{1-x}{1+x}\right)}\ dx}_{\frac{1-x}{1+x}=y}+\underbrace{\frac12\int_0^1\frac{\ln^4(1-x^2)}{x}\ dx}_{x^2=y}-\int_0^1\frac{\ln^4(1-x)}{x}\ dx$$
$$=\int_0^1\frac{\ln^4x}{1-x^2}+\frac14\int_0^1\frac{\ln^4(1-x)}{x}\ dx-\int_0^1\frac{\ln^4(1-x)}{x}\ dx$$ $$=\frac12\int_0^1\frac{\ln^4x}{1-x}+\frac12\int_0^1\frac{\ln^4x}{1+x}-\frac34\underbrace{\int_0^1\frac{\ln^4(1-x)}{x}\ dx}_{1-x=y}$$ $$=\frac12\int_0^1\frac{\ln^4x}{1+x}\ dx+\frac14\int_0^1\frac{\ln^4x}{1-x}\ dx=\frac12\left(\frac{45}{2}\zeta(5)\right)+\frac14(24\zeta(5))=\frac{21}4\zeta(5)$$
On the other hand, $\quad\displaystyle I=\underbrace{\int_0^1\frac{\ln^4(1+x)}{x}\ dx}_{I_1}+6\int_0^1\frac{\ln^2(1-x)\ln^2(1+x)}{x}\ dx$
Using $\ln^2(1+x)=2\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)x^n\ $ for the second integral, we get
\begin{align} I&=I_1+12\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\int_0^1x^{n-1}\ln^2(1-x)\ dx\\ I&=I_1+12\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\left(\frac{H_n^2+H_n^{(2)}}{n}\right)\\ I&=I_1+12\sum_{n=1}^\infty(-1)^n\left(\frac{H_n^3+H_nH_n^{(2)}}{n^2}\right)-12\sum_{n=1}^\infty(-1)^n\left(\frac{H_n^2+H_n^{(2)}}{n^3}\right)\tag{2} \end{align} From $(1)$ and $(2)$, we get
$$\boxed{\small{R_1=\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}+\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}=\frac{7}{16}\zeta(5)+\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^3}+\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^3}-\frac{1}{12}I_1}}$$
and the first relation is established.
To get the second relation, we need to use the sterling number formula ( check here) $$ \frac{\ln^k(1-x)}{k!}=\sum_{n=k}^\infty(-1)^k \begin{bmatrix} n \\ k \end{bmatrix}\frac{x^n}{n!}$$ letting $k=4$ and using $\displaystyle\begin{bmatrix} n \\ 4 \end{bmatrix}=\frac{1}{3!}(n-1)!\left[\left(H_{n-1}\right)^3-3H_{n-1}H_{n-1}^{(2)}+2H_{n-1}^{(3)}\right],$ we get $$\frac14\ln^4(1-x)=\sum_{n=1}^\infty\frac{x^{n+1}}{n+1}\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$
differentiate both sides with respect to $x$, we get $$-\frac{\ln^3(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$
Now replace $x$ with $-x$ then multiply both sides by $\frac{\ln x}{x}$ and integrate, we get $$-\sum_{n=1}^\infty(-1)^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)\int_0^1x^{n-1}\ln x\ dx=\int_0^1\frac{\ln^3(1+x)\ln x}{x(1+x)}\ dx$$ $$\sum_{n=1}^\infty \frac{(-1)^n}{n^2}\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)=\int_0^1\frac{\ln^3(1+x)\ln x}{x}\ dx-\underbrace{\int_0^1\frac{\ln^3(1+x)\ln x}{1+x}\ dx}_{IBP}$$ $$\sum_{n=1}^\infty \frac{(-1)^n}{n^2}\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)=\int_0^1\frac{\ln^3(1+x)\ln x}{x}\ dx+\frac14I_1$$ Rearranging the terms, we get $$\boxed{R_2=\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}-3\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}=\int_0^1\frac{\ln^3(1+x)\ln x}{x}-2\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}+\frac14I_1}$$ and the second relation is established.
Now we are ready to calculate the first sum. \begin{align} \sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}&=\frac{3R_1+R_2}{4}\\ &=\frac34\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^3}+\frac34\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^3}-\frac12\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}\\ &\quad+\frac14\int_0^1\frac{\ln x\ln^3(1+x)}{x}\ dx+\frac{21}{64}\zeta(5) \end{align} the closed form of the first and second sum can be found here and the closed form of the third sum is evaluated here. as for the integral, I evaluated it here. by combining these results, we get our closed form.
and the second sum. $$\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}=\frac{R_1-R_2}{4}$$ $$\small{=\frac14\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^3}+\frac14\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^3}+\frac12\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}-\frac14\int_0^1\frac{\ln x\ln^3(1+x)}{x}\ dx-\frac1{12}I_1+\frac{7}{64}\zeta(5)}$$ lets calculate $I_1$ and by setting $\frac1{1+x}=y$, we get \begin{align} I_1&=\int_0^1\frac{\ln^4(1+x)}{x}=\int_{1/2}^1\frac{\ln^4x}{x}\ dx+\int_{1/2}^1\frac{\ln^4x}{1-x}\ dx\\ &=\frac15\ln^52+\sum_{n=1}^\infty\int_{1/2}^1 x^{n-1}\ln^4x\ dx\\ &=\frac15\ln^52+\sum_{n=1}^\infty\left(\frac{24}{n^5}-\frac{24}{n^52^n}-\frac{24\ln2}{n^42^n}-\frac{12\ln^22}{n^32^n}-\frac{4\ln^32}{n^22^n}-\frac{\ln^42}{n2^n}\right)\\ &=4\ln^32\zeta(2)-\frac{21}2\ln^22\zeta(3)+24\zeta(5)-\frac45\ln^52-24\ln2\operatorname{Li}_4\left(\frac12\right)-24\operatorname{Li}_5\left(\frac12\right) \end{align} by combining the result of $I_1$ along with the results we used in our first sum, we get the closed form of the second sum.
UPDATE:
The identity used above:
$$-\frac{\ln^3(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$
can also be proved this way.
We will look into the integral \begin{align*} I = &\int_{0}^{1}\arcsin^4 x\frac{ \ln x}{\sqrt{1-x^2}}\ \mathrm dx \end{align*} taking the @nospoon's novel approach presented here. Using the MacLaurin series of $\arcsin^4 x$ $$ \arcsin^4 x =\frac 3 2 \sum_{n=1}^\infty \frac{4^{n}H_{n-1}^{(2)}}{n^2{2n \choose n}}x^{2n} $$ and the fact that $$ \small\operatorname{B}(n+\tfrac 1 2,\tfrac 1 2) = \int_0^1 x^{n-1/2}(1-x)^{-1/2}\ \mathrm dx = 2\int_0^{\frac\pi 2} \sin^{2n}\theta\ \mathrm d\theta = \frac{\pi}{4^n}{2n \choose n},\tag{$\small x\mapsto \sin^2\theta$} $$ \begin{align*}\small \psi(n+\tfrac 12 ) -\psi(n+1) =&\small \sum_{k=1}^\infty \frac 1{\scriptsize k+n} - \frac 1{\scriptsize k+n-\tfrac 1 2} \\ =&\small\sum_{k=1}^\infty \left(\frac 1{\scriptsize k} - \frac 1{\scriptsize k-\tfrac 1 2}\right)-\sum_{k=1}^n\frac 1 {\scriptsize k} + \sum_{k=1}^n\frac 1{\scriptsize k-\tfrac 1 2}\\ =&\small-2\ln 2 -H_n +2(H_{2n}-\tfrac 1 2H_n)\\ =&\small 2(H_{2n}-H_n-\ln 2), \end{align*} \begin{align*} \Longrightarrow \ {\int_{ 0}^{1 }x^{2n}\frac{ \ln x}{\sqrt{1-x^2}}\ \mathrm dx} = & \frac 1 4\int_{0 }^{1 }x^{n-1/2} { \ln x \over \sqrt{1-x}}\ \mathrm dx\tag{$\small x^2\mapsto x$}\\ =& \frac 1 4 \left[\frac{\partial }{\partial x}\operatorname{B}(x,y) \right]_{x=n+1/2,y=1/2}\\ =&\frac 1 4\Big[ \operatorname{B}(x,y)\big[\psi(x) -\psi(x+y) \big]\Big]_{x=n+1/2,y=1/2}\\ =& \frac 1 4 \operatorname{B}(n+\tfrac 1 2,\tfrac 1 2)\big[\psi(n+\tfrac 12 ) -\psi(n+1) \big]\\ =& \frac{\pi}2\frac{{2n \choose n}}{4^{n}} \left(H_{2n} - H_n -\ln 2\right), \end{align*} where $\operatorname{B}(x,y)$ and $\psi(x)$ are the Beta and digamma function, respectively, we have \begin{align*} I = &\frac 3 2\sum_{n=1}^\infty \frac{4^{n}H_{n-1}^{(2)}}{n^2{2n \choose n}}\int_{0}^{1}x^{2n}\frac{ \ln x}{\sqrt{1-x^2}}\ \mathrm dx \\ =&\frac {3\pi}4 \sum_{n=1}^\infty \frac{H^{(2)}_{n-1}}{n^2}\left(H_{2n} - H_n -\ln 2\right) \\ =&\frac {3\pi}4\sum_{n=1}^\infty \frac{H^{(2)}_{n-1}H_{2n}}{n^2}-\frac {3\pi}4\underbrace{\sum_{n=1}^\infty \frac{H^{(2)}_{n-1}H_{n}}{n^2}}_{=-2\zeta(5) +2\zeta(2)\zeta(3)}-\frac {3\pi\ln 2}4\underbrace{\sum_{n=1}^\infty \frac{H^{(2)}_{n-1}}{n^2}}_{=\frac{3}4 \zeta(4)}\\ =&\frac{3\pi}{4} \sum_{n=1}^\infty \frac{H^{(2)}_{n}H_{2n}}{n^2} -\frac{3\pi}4\underbrace{\sum_{n=1}^\infty \frac{H_{2n}}{n^4}}_{=\frac{37}{4}\zeta(5)-4\zeta(2)\zeta(3)} +\frac{3\pi}2 \zeta(5) -\frac{\pi^3}4\zeta(3) -\frac{\pi^5\ln 2}{160}\\ =&\boxed{3\pi S -\frac{87\pi}{16} \zeta(5) +\frac{\pi^3}{4}\zeta(3) -\frac{\pi^5\ln 2}{160}} \end{align*} where $S = \sum_{n=1}^\infty \frac{H_{2n}H^{(2)}_{n}}{4n^2}$ is the sum in question, and the known values of several Euler sums $$ \sum_{n=1}^\infty \frac{H^{(2)}_{n-1}H_{n}}{n^2}=-2\zeta(5) +2\zeta(2)\zeta(3),\tag{1} $$ $$\sum_{n=1}^\infty \frac{H^{(2)}_{n}}{n^2}=\frac{7}4 \zeta(4),\tag{2} $$ \begin{align*}\sum_{n=1}^\infty \frac{H_{2n}}{n^4} =& 8\sum_{n=1}^\infty \frac{H_{n}}{n^4}-8\sum_{n=1}^\infty \frac{(-1)^{n-1} H_{n}}{n^4}\\ =&8\big(3\zeta(5)-\zeta(2)\zeta(3)\big)-8\left(\frac{59}{32}\zeta(5)-\frac 1 2\zeta(2)\zeta(3)\right)\\ =&\frac{37}4\zeta(5) - 4\zeta(2)\zeta(3)\tag{3} \end{align*} are used.
Note: $(1)$ is in @nospoon's answer here, $(2)$ can be found here, and for $(3)$ you can see Euler's formula and here.
Evaluation of $I$: By making substitution $x = \sin \theta$ and using the Fourier series of $$ \ln (\sin\theta) = -\ln 2 -\sum_{k=1}^\infty \frac{ \cos(2k \theta)}{k}, $$ we get \begin{align*} I =& \int_{0}^{\frac\pi 2} \theta^4 \ln(\sin\theta)\ \mathrm d\theta\\ =&\int_{0}^{\frac\pi 2} \theta^4\left(-\ln 2 -\sum_{k=1}^\infty \frac{ \cos(2k \theta)}{k}\right)\ \mathrm d\theta\\ =& -\ln 2\int_0^{\frac \pi 2}\theta^4\ \mathrm d\theta-\sum_{k=1}^\infty \frac{1}{k}\underbrace{\int_{0}^{\frac\pi 2}\theta^4 \cos(2k \theta) \ \mathrm d\theta}_{\text{IBP}\times 4}\\ =& -\frac{\pi^5\ln 2}{160}-\sum_{k=1}^\infty \frac{1}{k}\cdot\left(-\frac{\pi^3}{8}\frac{(-1)^{k-1}}{k^2} +\frac{3\pi}{4}\frac{(-1)^{k-1}}{k^4}\right)\\ =&-\frac{\pi^5\ln 2}{160}+\frac{\pi^3}8\underbrace{\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^3}}_{=\frac 3 4 \zeta(3)} - \frac{3\pi}4\underbrace{\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^5}}_{=\frac{15}{16}\zeta(5)}\\ =&\boxed{-\frac{\pi^5\ln 2}{160}+\frac{3\pi^3}{32}\zeta(3) -\frac{45\pi}{64}\zeta(5).} \end{align*}
Combining these, we get the equation $$ 3\pi S-\frac{87\pi}{16} \zeta(5) +\frac{\pi^3}{4}\zeta(3) -\frac{\pi^5\ln 2}{160}=-\frac{\pi^5\ln 2}{160} +\frac{3\pi^3}{32}\zeta(3)-\frac{45\pi}{64}\zeta(5), $$hence it follows $$ \boxed{S = \frac{101}{64}\zeta(5) -\frac{5\pi^2}{96}\zeta(3).} $$
Addendum: By considering MacLaurin series of \begin{align*} \ln(1-x)\ln(1+x) =&-\sum_{k=1}^\infty \left(\frac{ H_{2k}}k-\frac{H_k}{k} + \frac1{2k^2}\right)x^{2k} \end{align*} and \begin{align*} \frac{H_k}{k^2} + \frac{H_k^{(2)}}{k} -\frac{\zeta(2)}{k} =& \frac{\partial }{\partial k}\left[-\frac{H_k}{k}\right]\\ =& \int_0^1 x^{k-1}\ln x\ln(1-x)\ \mathrm dx\\ =&4\int_0^1 x^{2k-1}\ln x \ln(1-x^2)\ \mathrm dx \end{align*} we have that \begin{align*} &\int_{0}^{1}\ln(1-x)\ln(1+x) \frac{\ln x\ln(1-x^2)}x \ \mathrm dx \\&=-\sum_{k=1}^\infty \left(\frac{ H_{2k}}k-\frac{H_k}{k} + \frac1{2k^2}\right)\int_{0}^{1}x^{2k-1} \ln x \ln(1-x^2)\ \mathrm dx \\ &=-\frac 1 4\sum_{k=1}^\infty \left(\frac{ H_{2k}}k-\frac{H_k}{k} + \frac1{2k^2}\right)\left(\frac{H_k}{k^2} + \frac{H_k^{(2)}}{k} -\frac{\zeta(2)}{k}\right). \end{align*} The integral can be attacked by considering algebraic identity $$ ab(a+b) = \frac 1 3 (a+b)^3 - \frac {a^3}3 -\frac{b^3}3 $$ with $a=\ln(1-x)$ and $b=\ln(1+x)$, and extant results.
For the sum, after expanding the summand, the only tricky part is $$ \sum_{k=1}^\infty\frac{H_{2k}H_k}{k^3}, $$ which can be found here. Then, the sum $\sum_{k=1}^\infty \frac{H_{2k}H_k^{(2)}}{4k^2}$ can be evaluated by solving the equation obtained.
Best Answer
By coincidence, I asked virtually the same question. I managed to answer it myself which is,
$$F_a(z) = \sum_{n=1}^{\infty}\frac{H_n}{n^a}z^n= S_{a-1,2}(z) + \rm{Li}_{a+1}(z)$$
with polylogarithm $\rm{Li}_{a+1}(z)$ and Nielsen generalized polylogarithm $S_{n,p}(z)$. The situation then is subsumed by the Nielsen polylogs.
There are formulas for general $z$ when $a=1,2$ (as you mentioned) as well as for $a=3$, though I am not sure if there is for higher.