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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.

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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.

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  • $\begingroup$ Ahh ,how can i got that? $\endgroup$ – Almaa Dec 5 '19 at 10:18
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    $\begingroup$ See the note I added to my answer. @Almaa $\endgroup$ – Kavi Rama Murthy Dec 5 '19 at 10:19
  • $\begingroup$ Got it! Thanks , i would ask that after multiplying the two coulmns we get $\sum_{n>=2}$.. in the cofficient (the 2nd one which you wrote)? $\endgroup$ – Almaa Dec 5 '19 at 10:34

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