Finding the cofficient of a generating function.

Question

What is the cofficient of $n^n$ in the generating function $(\log(1-x))^2$? First i took the taylor coulmn of $\log(1-x)$ and multiple it with itself.then i used the coushi formula. I got that the cofficient is $\sum_{j=0}{n}\frac{1}{(j+1)^2}$ , but im not sure that it is true.

Answer 1

The correct answer is $\sum_{\{j,k \geq 1, j+k=n\}} \frac 1 {jk}$ or $ \sum\limits_{j=1}^{n-1} \frac 1 {j(n-j)}$.

Note: when you multiply two power series you should use different symbols for the summation indices.