Find $[z^N]$ for $\frac{1}{1-z} \left(\ln \frac{1}{1-z}\right)^2$ Find $[z^N]$ for $\frac{1}{1-z} \left(\ln \frac{1}{1-z}\right)^2$. Here generalized harmonic numbers should be used. 
$$H^{(2)}_n = 1^2 + \frac{1}{2}^2 + \dots + \frac{1}{N}^2.$$
For now, I was able to find $[z^N]$ for $\left(\ln \frac{1}{1-z}\right)^2$ which is the following convolution for $N \ge 2$:
$$\sum_{1\le k \le N-1} \frac{1}{N-k}\frac{1}{k}$$
Next step would be partial sum but I don't see how all this leads me to generalized harmonic numbers.
 A: From 
$$-\log(1-z)=\sum_{n\geq 1}\frac{z^n}{n}$$
it follows that
$$\frac{-\log(1-z)}{1-z} = \sum_{n\geq 1} H_n z^n $$
and by applying $\int_{0}^{x}\ldots dx$ to both sides
$$ \log^2(1-x) = \sum_{n\geq 1}\frac{2H_n}{n+1}x^{n+1} = \sum_{n\geq 2}\frac{2H_{n-1}}{n} x^n. $$
It follows that for any $n\geq 2$ we have
$$ [z^n]\frac{\log^2(1-z)}{1-z} = 2\sum_{k=1}^{n}\frac{H_{k}-\frac{1}{k}}{k}=2\sum_{k=1}^{n}\frac{H_k}{k}-2 H_{n}^{(2)}. $$
On the other hand, by symmetry,
$$ \sum_{k=1}^{n}\frac{H_k}{k}=\sum_{1\leq j\leq k}\frac{1}{jk}=\frac{H_n^2+H_{n}^{(2)}}{2}. $$
It follows that
$$ \frac{\log^2(1-z)}{1-z}=\sum_{n\geq 1}\color{red}{\left(H_n^2-H_n^{(2)}\right)}z^n. $$
A: Using the generalization:

$$\sum_{n=1}^\infty a_nx^n=\frac1{1-x}\sum_{n=1}^\infty (a_n-a_{n-1})x^n,\quad a_{0}=0$$

Let $a_{n}=H_n^2$ to have
\begin{align}
\sum_{n=1}^\infty H_n^2x^n&=\frac1{1-x}\sum_{n=1}^\infty \left(H_n^2-H_{n-1}^2\right)x^n\\
&=\frac1{1-x}\sum_{n=1}^\infty \left(\frac{2H_n}{n}-\frac1{n^2}\right)x^n\\
&=\frac1{1-x}\cdot 2\sum_{n=1}^\infty\frac{H_n}{n}x^n-\frac{\operatorname{Li}_2(x)}{1-x}\\
&=\frac1{1-x}\cdot 2\left(\operatorname{Li}_2(x)+\frac12\ln^2(1-x)\right)-\frac{\operatorname{Li}_2(x)}{1-x}\\
&=\frac{\ln^2(1-x)}{1-x}+\frac{\operatorname{Li}_2(x)}{1-x}\\
&=\frac{\ln^2(1-x)}{1-x}+\sum_{n=1}^\infty H_n^{(2)}x^n
\end{align}
Thus 

$$\frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^\infty (H_n^2-H_n^{(2)})x^{n}$$


The proof of the generalization together with other identities can be found here.
