<s>[INST] 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.
 [/INST] 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.
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