<s>[INST] Cauchy Product starting from $1$ The definition of the Cauchy product from Wikipedia is defined as
$$\left(\sum_{i=0}^\infty a_i\right)\left(\sum_{j=0}^\infty b_j\right)
=\sum_{k=0}^\infty\sum_{\ell=0}^ka_\ell b_{k-\ell}$$

Question: Does this apply if the summation runs from $1$ instead of $0$ for both $i,j$?

In particular, can it be said that
$$(\ln 2)^ 2=\left(\sum_{r=1}^\infty \frac {(-1)^{r+1}}r\right)^2
=\sum_{k=1}^\infty\sum_{\ell=1}^k \frac {(-1)^{\ell+1}}\ell\cdot \frac {(-1)^{k-\ell+1}}{k-\ell}\\
=\sum_{k=1}^\infty\sum_{\ell=1}^k \frac {(-1)^k}{\ell (k-\ell)} \text{?}$$

If not how can the Cauchy product be used to express $(\ln 2)^2$ as a double summation?

(NB - A related question is found here).
 [/INST] What you need is $\displaystyle \sum_{i=1}^\infty \sum_{j=1}^\infty a_{i,j} = \sum_{k=1}^\infty \sum_{\ell=1}^k a_{\ell,\,k+1-\ell}.$
$$
\begin{array}{c|ccccccccc}
& 1 & 2 & 3 & 4 & 5 & \longrightarrow j  \\
\hline
1 & a_{11} & a_{12} & a_{13} & a_{14} & \cdots \\
2 & a_{21} & a_{22} & a_{23} \\
3 & a_{31} & a_{32} \\
4 & a_{41} \\
5 & \vdots \\
\downarrow \\
i
\end{array}
$$
Here are the terms in which $k=4:$
$$
\begin{array}{c|ccccccccc}
& 1 & 2 & 3 & 4 & 5 & \longrightarrow i  \\
\hline
1 & & & & a_{14} \\
2 & & & a_{23} \\
3 & & a_{32} \\
4 & a_{41} \\
5 \\
\downarrow \\
j
\end{array}
$$
When $k=4,$ you have
$$
a_{14} + a_{23} + a_{32} + a_{41}.
$$
In each case, the sum of the two indices is $5,$ not $4\text{:}$ $\quad 1+4,\quad 2+3,\quad 3+2,\quad 4+1.$
So the sum $a_{14} + a_{23} + a_{32} + a_{41}$ is $\displaystyle \sum_{\ell=1}^4 a_{\ell,\,5-\ell}.$
Let us contrast that with the situation where you start with $0$ rather than with $1.$
$$
\begin{array}{c|ccccccccccc}
& 0 & 1 & 2 & 3 & 4 & 5 & \longrightarrow i  \\
\hline
0 & & & & & a_{04} \\
1 & & & & a_{13} \\
2 & & & a_{22} \\
3 & & a_{31} \\
4 & a_{40} \\
5 \\
\downarrow \\
j
\end{array}
$$
Here, in each case, the sum of the two indices is $4\text{:} \quad 0+4,\quad 1+3, \quad 2+2, \quad 3+1, \quad 4+0.$
So you have $\displaystyle \sum_{\ell=0}^4 a_{\ell,\,4-\ell}.$
</s>