Cauchy Product starting from $1$

Question

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).

Answer 1

I don't see anything wrong with your calculation here. Moreover, in answer to your question about it starting from 1, it is fine. Care will need to be taken just in case, especially if the sums started from different values to each other. The Cauchy product starting from 1 just corresponds to the Cauchy product starting from 0, but where the terms given for the "0-th" entry is 0.

Answer 2

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 $\displaystyle \sum_{\ell=0}^4 a_{\ell,\,4-\ell}.$

Answer 3

If $$a_0=b_0=0,$$ then \begin{align*}\left(\sum_{i=1}^\infty a_i\right)\left(\sum_{j=1}^\infty b_j\right)& =\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}\\ &= a_0b_0+(a_0b_1+a_1b_0)+\sum_{k=2}^\infty\sum_{\ell=0}^{k}a_\ell b_{k-\ell}\\ &= \sum_{k=2}^\infty\Big(a_0b_k+a_kb_0+\sum_{\ell=1}^{k-1}a_\ell b_{k-\ell}\Big)\\ &=\sum_{k=2}^\infty\sum_{\ell=1}^{k-1}a_\ell b_{k-\ell}\\ &=\sum_{k=1}^\infty\sum_{\ell=1}^{k}a_\ell b_{k+1-\ell}\\ \end{align*}