how to do cauchy product of sums For example, how to do cauchy product of $$\ln(1-x)*\ln (1-x)=\ln ^2 (1-x)$$
should it go like this?
$$\ln(1-x)=-\sum_{k=1}^{k=\infty}\frac{x^{k}}{k}$$
$$\ln ^2 (1-x)=\sum_{n=1}^{n=\infty}x^{n}\sum_{k=1}^{k=n}\frac{1}{k(n-k)}$$
Here , when n=k its 1/0 , how to fix this?
Also, for example, how ot find $$\arctan ^2(x)$$. thank you.
 A: Cauchy product of $\sum_0^\infty a_n x^n $ with $\sum_0^\infty b_n x^n $ is $\sum_0^\infty c_n x^n $ where $c_n=\sum_{k+l=n} a_k b_l$. To apply this formula you should "pad the series" with a zero at the zeroth term:
$$\ln(1-x) = \sum_{n=0}^\infty a_n x^n, \quad a_n= \begin{cases} 0 & n=0 \\ -1/n & n>0\end{cases}.$$
Then
$ \ln^2(1-x)=\sum_{n=0}^\infty c_n x^n$ with
$$ c_n = \sum_{k+l=n} a_k a_l=\sum_{\substack{k+l=n \\ k>0\text{ and }l>0}} a_k a_l + \underbrace{\sum_{\substack{k+l=n \\ k=0\text{ or }l=0}} a_k a_l}_{=0}=\sum_{k=1}^{n-1} \frac1{k(n-k)}$$
A: When doing the series multiplication we can perform the cauchy multiplication as usual, but we have to carefully consider the index limits. We obtain
\begin{align*}
\color{blue}{\ln^2(1-x)}&=\left(-\sum_{k=1}^\infty \frac{x^k}{k}\right)\tag{1}
\left(-\sum_{l=1}^\infty \frac{x^l}{l}\right)\\
&=\sum_{n=2}^\infty\left(\sum_{\substack{k+l=n\\k,l\geq 1}}\frac{1}{k}\,\frac{1}{l}\right)x^n\tag{2}\\
&\,\,\color{blue}{=\sum_{n=2}^\infty\left(\sum_{k=1}^{n-1}\frac{1}{k(n-k)}\right)x^n}\tag{3}\\
\end{align*}
Comment:

*

*In (1) we see the  indices $k,l$ of both series start with lower limit $1$, so that $x^2$ is the term with the smallest positive exponent.


*In (2) we collect the terms according to increasing powers of $n$. So, we consider indices $k+l=n$, but we also have to respect that $k,l\geq 1$ by explicitly noting it and we also start with lower index limit $n=2$.


*In (3) we replace $l$ by $n-k$. Since both, $l,k\geq 1$ we have lower index limit $k=1$ and upper limit $k=n-1$.
