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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).

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    $\begingroup$ I'd rather call it $(\ln 2)^2$ because sometimes the notation $\ln^2 2$ is used to mean $\ln\ln2. \qquad$ $\endgroup$ Oct 23 '17 at 17:12
  • $\begingroup$ off question. Does Cauchy product you are using apply to such series? $\endgroup$
    – Guy Fsone
    Oct 23 '17 at 19:28
  • $\begingroup$ Because I know at least one of the series must be absolutely convergent $\endgroup$
    – Guy Fsone
    Oct 23 '17 at 19:32
  • $\begingroup$ en.wikipedia.org/wiki/Cauchy_product $\endgroup$
    – Guy Fsone
    Oct 23 '17 at 19:32
  • $\begingroup$ Need $k-l+1$ instead of $k-l$ otherwise have problems when$k=l$. $\endgroup$ Oct 23 '17 at 23:50
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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 $\displaystyle \sum_{\ell=0}^4 a_{\ell,\,4-\ell}.$

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  • $\begingroup$ Why have you isolated the case when $k=4$ here? $\endgroup$
    – user383264
    Oct 23 '17 at 18:12
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    $\begingroup$ @CharleyScotford : Because it appeared to be the simplest example that's not too simple to understand. $\endgroup$ Oct 23 '17 at 18:14
  • $\begingroup$ +1 I really like the way you present the thing. Does it also means that we can rearrange the sums with any enumeration of $\Bbb N^2$ or it only works for the diagonal enumeration? $\endgroup$
    – Surb
    Oct 23 '17 at 18:51
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    $\begingroup$ @Surb : I'm not sure I understand your question. Certainly this way of arranging the sum is one way to enumerate the members of $\mathbb N^2.$ But what does it mean to say it "works"? If it means the sum is unchanged by the rearrangement, then that is true if you have absolute convergence, and that doesn't depend on the way in which the sum was originally arranged (in this case with two indices $i$ and $j$). Which rearrangements alter the sum in the absence of absolute convergence is a more complicated question, not addressed here. $\endgroup$ Oct 23 '17 at 19:20
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    $\begingroup$ Yes. Perhaps a separate question could address that at greater length. $\endgroup$ Oct 23 '17 at 23:33
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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*}

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    $\begingroup$ Thanks for your answer. (+1) $\endgroup$ Oct 24 '17 at 16:13
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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.

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    $\begingroup$ I disagree. See my posted answer. $\endgroup$ Oct 23 '17 at 17:43
  • $\begingroup$ Thanks for your answer. However what about the case where the $0$-th entry is not defined, like in the case of the expansion for $ln (1+x)$? $\endgroup$ Oct 24 '17 at 16:15

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