TITLE: A conjectured result for $\sum_{n=1}^\infty\frac{(-1)^n\,H_{n/5}}n$
QUESTION [19 upvotes]: Let $H_q$ denote harmonic numbers (generalized to a non-integer index $q$):
$$H_q=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+q}\right)=\int_0^1\frac{1-x^q}{1-x}dx=\gamma+\psi(q+1),\tag1$$
where $\psi(z)=\Gamma'(z)/\Gamma(z)$ is the digamma function.
My goal is to evaluate the following series:
$$\mathcal S_m=\sum_{n=1}^\infty\frac{(-1)^n\,H_{n/m}}n.\tag2$$
Using the integral representation from $(1)$ we can get equivalent integral forms:
$$\mathcal S_m=\int_0^1\frac{\ln(1+\sqrt[m]x)-\ln2}{1-x}\,dx=m\int_0^1\frac{\ln(1+z)-\ln2}{1-z^m}\,z^{m-1}dz.\tag3$$
Here are some simple cases:
$$\begin{align}&\mathcal S_1=\frac{\ln^22}2-\frac{\pi^2}{12}\hspace{7.7em}\color{maroon}{\mathcal S_2=\ln^22-\frac{\pi^2}{12}}\\\\&\color{blue}{\mathcal S_3=\frac{3\ln^22}2-\frac{\pi^2}9+\frac12\,\operatorname{Li}_2\!\left(\tfrac14\right)}\hspace{2em}\color{green}{\mathcal S_4=\frac{7\ln^22}4-\frac{5\pi^2}{48}}\end{align}\tag4$$
For $\mathcal S_5$ the integral can be found using Mathematica (there is even a closed-form antiderivative, so it should be possible in principle to prove it by differentiation), but the result takes tens of thousands characters to write down (you can see it here), and Mathematica cannot do much simplification on it (here is a simplified result).
But I was able to conjecture a much simpler closed form that fits numerically with a high precision:

$$\mathcal S_5\stackrel{\color{gray}?}=\frac{\ln^22}2-\frac{\ln^25}4+\ln2\cdot\ln5-\frac12\,\operatorname{Li}_2\!\left(\tfrac15\right)-\operatorname{Li}_2\!\left(\frac{\sqrt5-1}2\right)\tag{$\diamond$}$$

I hope there is a way to prove this result manually without going through huge intermediate expressions, but so far I have not found it.

REPLY [3 votes]: General solution 
I have found a systematic path to a general solution. 
It turns out the the generalized problem is easier to solve and that, as a by-product, it reveals clearly the structure of the problem.
So for natural $m$ consider the sum
$$s_{m} = \sum_{n=1}^\infty\frac{(-1)^n\,H_{n/m}}n\tag{1}$$
Doing the $k$-sum after replacing $H_{z}=\int_{0}^{1} \frac{1-x^z}{1-x}\,dx, z\to\frac{k}{m}$ we can write
$$\sum _{k=1}^{\infty } \frac{(-1)^k \left(1-x^{k/m}\right)}{k (1-x)}=\int_0^1 \frac{-\log (2)+\log \left(1+x^{1/m}\right)}{1-x}\,dx
\\\overset{x\to t^m}=-\int_0^1 m t^{m-1}\left(\frac{\log(2)-\log(1+t)}{1-t^m}   \right)\,dt
\\
=\log(2)\log(1-t^m) + m \int t^{m-1}\left(\frac{\log(1+t)}{1-t^m}\right)\,dt\tag{2}$$
Here in the last line we have switches to indeinite integrals to avoid spurious divergencies.
Integration by parts of the last integral with 
$$u = \int m \frac{t^{m-1}}{1-t^m}\,dt = - \log(1-t^m), v = \log(1+t)\tag{3}$$ 
gives
$$m \int t^{m-1}\left(\frac{\log(1+t)}{1-t^m}\right)\,dt\\
= - \log(1-t^m)\log(1+t) + \int \frac{\log(1-t^m)}{1+t}\,dt\tag{4}$$
Collecting what we have up to now in 
$$f_{m}(t) = - \log(1-t^m)\log(\frac{1+t}{2}) + \int \frac{\log(1-t^m)}{1+t}\,dt \tag{5}$$
so that
$$s_m = f_{m}(t\to 1)-f_{m}(t\to0)= \int_0^1 \frac{\log(1-t^m)}{1+t}\,dt\tag{6}$$
because the product of the logs vanishes at both ends. For what follows it will be convenient technically to proceed with the indefinite version $f_m(t)$.
The integral can be calculated. Writing 
$$1-t^m = \prod_{k=1}^{m} (\alpha_{m,k} - t)$$
where the roots are
$$\alpha_{m,k} = \exp(2 i \pi \frac{k}{m}), k=1..m\tag{7}$$
In what follows we will drop the index $m$ in $\alpha$ for simplicity.
Now the log can be written as
$$\log(1-t^m) = \log(\prod_{k=1}^{m}(\alpha_{m,k}-t))= \sum_{k=1}^{m}\log(\alpha_{m,k}-t)\tag{8}$$
and the integral boils down to terms of the form
$$\int \frac{\log (\alpha -t)}{t+1} \, dt =\operatorname{Li}_2\left(\frac{\alpha -t}{\alpha +1}\right)+\log \left(\frac{t+1}{\alpha +1}\right) \log (\alpha -t)\tag{9}$$
Now observing that we can write
$$- \log(1-t^m)\log\left(\frac{1+t}{2}\right)= -\log\left(\frac{1+t}{2}\right)\sum_{k=1}^{m} \log(\alpha_k - t)\tag{10}$$
the indefinite integral $(5)$ can be written as
$$f_{m}(t) =  \sum_{k=1}^m \left(\operatorname{Li}_2\left(\frac{\alpha_{k} -t}{\alpha_{k} +1}\right)
\\
+\log \left(\frac{t+1}{\alpha_{k} +1}\right) \log (\alpha_{k} -t)-\log\left(\frac{1+t}{2}\right) \log(\alpha_k - t)\right)\tag{11}$$
To make this expression definite (and finite) we follow $(6)$ and take the value at $t=1$ and subtract its value at $t=0$
$$\phi_{m}(t) = f_{m}(t)- f_{m}(0)$$
This results in a sum over
$$A_{k} = \operatorname{Li}_2\left(\frac{\alpha_{k}-1}{\alpha_{k}+1}\right)-\operatorname{Li}_2\left(\frac{\alpha_{k}}{\alpha_{k}+1}\right)-\log\left(1-\frac{1}{\alpha_{k}}\right)\log\left(\frac{1+\alpha_{k}}{2}\right)\tag{12}$$
There's a still little technical difficulty to circumvent: it turns out that inserting the value of the index $k$ as an integer leads to a divergent behaviour of some $A_{k}$. 
Hence we understand $A_{k}$ as the limit from below $A_{k}=\lim_{\epsilon\to +0} \,A_{k-\epsilon}$ 
Now we are ready to write down the closed expression for the sum $(1)$
$$s_{m} = \sum_{k=1}^{m} A_{k}\tag{13}$$
The general structure of the solution is clear: the sum is composed of quadratic $\log$s and linear $\operatorname{Li}_2$s. No higher polylogs appear.
Remark 1: The main remaining task would be a possible simplification of the polylog function $\operatorname{Li}_2(z)$ of a composite and complex arguments.
Remark 2: It seems possible to do a similar analysis for any rational $m=\frac{p}{q}$.
Remark 3: as we have encountered only antiderivatives before the final limits, we can apply the same reasoning to find the generating function 
$$g_{m}(z) = \sum_{k=1}^{\infty}\frac{z^k}{k} H_{\frac{k}{m}}$$
Evaluation of specific cases
Now we turn to the harvest. But I ask for some patience.
Original post
With the help of Mathematica I find a long expression which is numerically correct but needs plenty of simplification in order to become "culturally acceptable".
Doing the $k$-sum after replacing $H_{z}=\int_{0}^{1} \frac{1-x^z}{1-x}\,dx, z\to\frac{k}{5}$ gives the integral 
$$i= \int_0^1 \frac{\log (2)-\log \left(\sqrt[5]{x}+1\right)}{x-1} \, dx$$ 
Letting $x\to t^5$ the integrand of $dt$ becomes
$$\frac{5 t^4 \log \left(\frac{2}{t+1}\right)}{t^5-1}=\frac{5 t^4 \log (2)}{t^5-1}-\frac{5 t^4 \log (t+1)}{t^5-1}$$
To avoid divergences we continue with the indefinite integral.
The first integral is 
$$i_1= \int \frac{5 t^4 \log (2)}{t^5-1} = \log (2) \log \left(t^5-1\right)$$
and the second one is done by Mathematica returning a lengthy expression. 
The we form the sum $i=i_1 + i_2$ and calculate the limits $t\to0$ and $t\to1$ to find this horrendeous expression
$$i = -\text{Li}_2\left(\frac{\sqrt[5]{-1}}{-1+\sqrt[5]{-1}}\right)+\text{Li}_2\left(\frac{1+\sqrt[5]{-1}}{-1+\sqrt[5]{-1}}\right)-\text{Li}_2\left(\frac{(-1)^{2/5}}{1+(-1)^{2/5}}\right)+\text{Li}_2\left(\frac{-1+(-1)^{2/5}}{1+(-1)^{2/5}}\right)-\text{Li}_2\left(\frac{(-1)^{3/5}}{-1+(-1)^{3/5}}\right)+\text{Li}_2\left(\frac{1+(-1)^{3/5}}{-1+(-1)^{3/5}}\right)-\text{Li}_2\left(\frac{(-1)^{4/5}}{1+(-1)^{4/5}}\right)+\text{Li}_2\left(\frac{-1+(-1)^{4/5}}{1+(-1)^{4/5}}\right)+\frac{43 \pi ^2}{60}+\frac{\log ^2(2)}{2}+\log (5) \log (2)-\log \left(\sqrt[5]{-1}-1\right) \log \left(1+\sqrt[5]{-1}\right)-\log \left((-1)^{2/5}-1\right) \log \left(1+(-1)^{2/5}\right)-\log \left((-1)^{3/5}-1\right) \log \left(1+(-1)^{3/5}\right)-\log \left((-1)^{4/5}-1\right) \log \left(1+(-1)^{4/5}\right)+\frac{1}{5} i \pi  \left(\log \left(\sqrt[5]{-1}-1\right)+5 \log \left(1+\sqrt[5]{-1}\right)+2 \log \left(1+(-1)^{2/5}\right)+3 \log \left((-1)^{3/5}-1\right)+5 \log \left(1+(-1)^{3/5}\right)+4 \log \left(1+(-1)^{4/5}\right)\right)$$
The numeric values of which is $N(i) = -0.152666...$