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

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... 
