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Why multiplication of finite sums $(\sum_{i=0}^n a_i)(\sum_{i=0}^n b_i)=\sum_{i=0}^n (\sum_{j=0}^ia_jb_{i-j})$ (EDIT: This assumption was shown to be false) does not work in infinite case? I have constructed proof which shows it does but it must hase some flaw which I can not find. Here goes the proof:

By definition infinite series is just limit $\sum_{n=0}^\infty a_n = \lim_{n \to \infty } \sum_{i=0}^n a_i$. So using this definition, multiplication of finite sums and distributivity of $lim$:

$$\left(\sum_{n=0}^\infty a_n\right)\left(\sum_{i=0}^\infty b_i\right) = \lim_{n\to\infty}\sum_{i=0}^n a_i \cdot\lim_{n \to \infty } \sum_{i=0}^n b_i = \lim_{n \to \infty } \sum_{i=0}^n a_i \sum_{i=0}^n b_i $$ $$= \lim_{n \to \infty } \sum_{i=0}^n \sum_{j=0}^ia_jb_{i-j} = \sum_{i=0}^\infty \sum_{j=0}^ia_jb_{i-j}$$

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AFAIK it works for absolutely convergent (and that can be even relaxed a bit by not requiring all of them to be). – user2345215 Mar 9 '14 at 14:23
@user2345215 I know. Requirement is that at least one of the sum must be absolutely convergent. – Trismegistos Mar 9 '14 at 18:40
up vote 5 down vote accepted

This is certainly not true:

$$\left(\sum_{i=0}^n a_i\right)\left(\sum_{i=0}^nb_i\right) = \sum_{i=0}^n \sum_{j=0}^i a_jb_{i-j}$$

So it's not clear why you'd expect the limit as $n\to\infty$ is the same. If definitely requires a much more careful argument.

The actual equality is:

$$\left(\sum_{i=0}^n a_i\right)\left(\sum_{i=0}^nb_i\right) = \sum_{i=0}^n \sum_{j=0}^i a_jb_{i-j} + \sum_{i=0}^n\sum_{j=i}^n a_jb_{n+i-j}$$

You need that error term to approach zero, which is not at all obvious.

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Wikipedia article ( suggested that $\left(\sum_{i=0}^n a_i\right)\left(\sum_{i=0}^nb_i\right) = \sum_{i=0}^n \sum_{j=0}^i a_jb_{i-j}$ for finite sums as well as formal power series. – Trismegistos Mar 9 '14 at 15:21
Just work it out yourself - the right hand side does not have $a_nb_n$ in it. And nowhere does the Wikipedia article say what you think it says. Rather, for finite sums, you just treat it as an infinite sum where $a_k=0$ when $k>n$. You still need terms where $i+j>n$. @Trismegistos – Thomas Andrews Mar 9 '14 at 15:27

Just try with two terms. So, you have $$(\sum_{i=0}^1 a_i)(\sum_{i=0}^1 b_i)=a_0 b_0+a_1 b_0+a_0 b_1+a_1 b_1$$ $$\sum_{i=0}^1 \sum_{j=0}^1 a_jb_{i-j}=a_0 b_0+a_1 b_0+a_0 b_1$$

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As @Claude has stated for the simpler cases here is it for 4 elements in $a$ and in $b$. The correct sum is the sum over all elements of the ("outer"(?)) product $C$ of the two vectors $$ A^T \cdot B=C= \small \begin{array} {r|rrrr} & b_0 & b_1 & b_2 & b_3 \\ \hline a_0 & a_0b_0 & a_0b_1 & a_0b_2 & a_0b_3 \\ a_1 & a_1b_0 & a_1b_1 & a_1b_2 & a_1b_3 \\ a_2 & a_2b_0 & a_2b_1 & a_2b_2 & a_2b_3 \\ a_3 & a_3b_0 & a_3b_1 & a_3b_2 & a_3b_3 \end{array} $$ But your second expression which refers to the antidiagonals sums only the antidiagonals up to (and including) $a_3b_0 ... a_0 b_3$ but not the remaining antidiagonals.

For the infinite cases this can only be correct if the sum of the whole remaining lower triangle beyond $$ C^*= \small \begin{array} {r|rrrr} & b_0 & b_1 & b_2 & b_3 \\ \hline a_0& & & & \\ a_1 & & & & a_1b_3 \\ a_2& & & a_2b_2 & a_2b_3 \\ a_3& & a_3b_1 & a_3b_2 & a_3b_3 \end{array} $$ would be neglectable (converges to zero). But notice that its size expands also without bound when n tends to infinity...

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