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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).

Hypergeometricx
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3 Answers3

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

Michael Hardy
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  • Why have you isolated the case when $k=4$ here? –  Oct 23 '17 at 18:12
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    @CharleyScotford : Because it appeared to be the simplest example that's not too simple to understand. – Michael Hardy Oct 23 '17 at 18:14
  • +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? – Surb Oct 23 '17 at 18:51
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    @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. – Michael Hardy Oct 23 '17 at 19:20
  • @Surb : I'm glad you liked it. – Michael Hardy Oct 23 '17 at 19:20
  • Sorry for being unprecise (reading the wiki article before commenting would have helped....). Formally, assuming both $\sum_{k=0}^{\infty}a_k$ and $\sum_{k=0}^{\infty}b_k$ converge absolutely, is it then true that for every bijection $\phi \colon \Bbb N \to \Bbb N \times \Bbb N$ with $\phi(n)=(\phi_1(n),\phi_2(n))$, it holds $(\sum_{k=0}^\infty a_k)(\sum_{k=0}^\infty b_k)=\sum_{k=0}^{\infty} a_{\phi_1(k)}b_{\phi_2(k)}$? – Surb Oct 23 '17 at 19:55
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    Yes. Perhaps a separate question could address that at greater length. – Michael Hardy Oct 23 '17 at 23:33
  • Thanks for your answer. Very nice example. (+1) – Hypergeometricx Oct 24 '17 at 16:13
  • @hypergeometric : I'm glad it helped. – Michael Hardy Oct 24 '17 at 16:31
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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*}

Surb
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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.