Power series of $\log(1+x)\log(1-x)$ On the bottom of page 7 of the paper Logarithmic Integrals by Morshed (arXiv link) there is an "interesting generating function":
$$ \log(1+x) \log(1-x) = \sum_{n=1}^\infty \left(H_n - H_{2n} - \frac{1}{2}\right) \frac{x^{2n}}{n}$$
but it doesn't seem to be correct. The function on the left has the series
$$-x^2 - \frac{5}{12}x^4 - \frac{47}{180}x^6 - \ldots$$
while the function on the right is
$$-x^2 - \frac{13}{24}x^4 - \frac{67}{180} - \ldots$$
These do not match. Not knowing how to actually find the power series for $\log(1+x)\log(1-x)$, I decided instead to try to identify the function on the right, using the identity
$$\sum_{n=1}^\infty H_n x^n = \frac{-\log(1-x)}{1-x}$$
as a starting point.
In the end I was able to obtain
$$\sum_{n=1}^\infty \left(H_n - H_{2n} - \frac{1}{2}\right) \frac{x^{2n}}{n} = \log(1+x)\log(1-x) + \tfrac{1}{2}\left(\mathrm{Li}_2(x^2) + \log(1-x^2)\right).$$
But the power series of $\mathrm{Li}_2(x^2) + \log(1-x^2)$ is very simple (it's $\displaystyle \sum_{n=1}^\infty \left(\frac{1}{n^2} - \frac{1}{n}\right) x^{2n}$) and so
$$\log(1+x)\log(1-x) = \sum_{n=1}^\infty\left(H_n - H_{2n} - \frac{1}{2\color{red}{n}}\right) \frac{x^{2n}}{n} \tag{*}$$
which means the paper had a typo.
Question. How would you derive (*) without knowing it had anything to do with harmonic numbers?
I'm dissatisfied with my derivation because I started with a "guess" that was "close" (by virtue of it being a typo) and was able to go from there. I'd be especially interested in a "power series proof" (multiplying together the series for $\log(1-x) = -x - x^2/2 - x^3/3 - \ldots$ and $\log(1+x) = x - x^2/2 + x^3/3 - \ldots$ and simplifying the coefficients).
 A: 
We obtain
\begin{align*}
\color{blue}{\log}&\color{blue}{(1+x)\log(1-x)}\\
&=\left(\sum_{j=1}^\infty (-1)^{j+1}\frac{x^j}{j}\right)\left(-\sum_{k=1}^\infty\frac{x^k}{k}\right)\\
&=\sum_{n=2}^\infty\left(\sum_{{j+k=n}\atop{j,k\geq 1}}\frac{(-1)^j}{j}\,\frac{1}{k}\right)x^n\\
&=\sum_{n=2}^\infty\left(\sum_{j=1}^{n-1}\frac{(-1)^j}{j}\,\frac{1}{n-j}\right)x^n\\
&=\sum_{n=2}^\infty\left(\sum_{j=1}^{n-1}\frac{(-1)^j}{j}+\sum_{j=1}^{n-1}\frac{(-1)^j}{n-j}\right)\frac{x^n}{n}\tag{1}\\
&=\sum_{n=2}^\infty\left(\sum_{j=1}^{n-1}\frac{(-1)^j}{j}+\sum_{j=1}^{n-1}\frac{(-1)^{n-j}}{j}\right)\frac{x^n}{n}\tag{2}\\
&=\sum_{n=1}^\infty\left(\sum_{j=1}^{2n-1}\frac{(-1)^j}{j}\right)\frac{x^{2n}}{n}\tag{3}\\
&=\sum_{n=1}^\infty\left(-\sum_{j=1}^{n}\frac{1}{2j-1}+\sum_{j=1}^{n}\frac{1}{2j}-\frac{1}{2n}\right)\frac{x^{2n}}{n}\tag{4}\\
&=\sum_{n=1}^\infty\left(-H_{2n}+2\sum_{j=1}^{n}\frac{1}{2j}-\frac{1}{2n}\right)\frac{x^{2n}}{n}\tag{5}\\
&\,\,\color{blue}{=\sum_{n=1}^\infty\left(-H_{2n}+H_n-\frac{1}{2n}\right)\frac{x^{2n}}{n}}\\
\end{align*}
and the claim follows.

Comment:

*

*In (1) we use the identity $\frac{1}{j(n-j)}=\frac{1}{n}\left(\frac{1}{j}+\frac{1}{n-j}\right)$.


*In (2) we change the order of the right-hand inner sum $j\to n-j$.


*In (3) we cancel odd terms of $n$.


*In (4) we split the inner sum in odd and even part.


*In (5) we use the identity $H_{2n}=\sum_{j=1}^{2n}\frac{1}{j}=\sum_{j=1}^{n}\frac{1}{2j}+\sum_{j=1}^n\frac{1}{2j-1}$.
A: Replace $y$ by $-y$ in the generating function:
$$\frac{\ln(1+y)}{1-y}=\sum_{n=1}^\infty \overline{H}_ny^{n},$$
we get
$$\frac{\ln(1-y)}{1+y}=\sum_{n=1}^\infty (-1)^n\overline{H}_ny^{n}.$$
Subtract the two generating functions,
$$\frac{\ln(1-y)}{1+y}-\frac{\ln(1+y)}{1-y}=\sum_{n=1}^\infty (-1)^n\overline{H}_ny^{n}-\sum_{n=1}^\infty \overline{H}_ny^{n}.$$
Integrate both sides from $y=0$ to $x$ using:
$$\int_0^x \left(\frac{\ln(1-y)}{1+y}-\frac{\ln(1+y)}{1-y}\right)\mathrm{d}y=\ln(1-x)\ln(1+x),$$
we have
\begin{gather*}
\ln(1-x)\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^n\overline{H}_nx^{n+1}}{n+1}-\sum_{n=1}^\infty \frac{\overline{H}_nx^{n+1}}{n+1}\\
\left\{\text{use $\sum_{n=1}^\infty (-1)^n a_n-\sum_{n=1}^\infty  a_n=-2\sum_{n=1}^\infty a_{2n-1}$}\right\} \\
=-2\sum_{n=1}^\infty \frac{\overline{H}_{2n-1}}{2n}x^{2n}\\
\left\{\text{use $\overline{H}_{n-1}=\overline{H}_{n}+\frac{(-1)^n}{n}$}\right\}\\
=-\sum_{n=1}^\infty \frac{\overline{H}_{2n}+\frac{1}{2n}}{n}x^{2n},
\end{gather*}
and the proof follows on using $\overline{H}_{2n}=H_{2n}-H_{n}$.
