Calculus – Proof of Cauchy Product

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I'm searching for a proof of the Cauchy product that states:

If the series $\sum a_n$, $\sum b_n$ and $\sum c_n$ converge to $A$, $B$ and $C$, and $c_n = a_0b_n+\cdots a_nb_0$ then $C = AB$

All the proofs I found start with the assumption

$$c_n = \sum_{k=0}^n a_kb_{n-k}$$ and then use some other techniques to prove that the sum $$C_n =\sum_{k=0}^n c_k$$ converges to $AB$

I can verify the proof but I have no intuition on why the $c_n$ term is like it is. Maybe there's a particular case of a simple sum where I can note this pattern?

For example, I've expanded a finite product in mathematica:

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Could someone at least show me a product of two finite sums such that I can see this pattern $a_k b_{n-k}$ emerging from the product? I can't see this pattern in the expansion. Maybe that's now how it Works but I think I should note something in the finite case.

Best Answer

Here's an idea. Arrange the terms in a table like this: $$ \begin{array}{ cccccc } \vdots & \vdots & \vdots & \vdots & \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.} \\ a_3b_0 & a_3b_1 & a_3b_2 & a_3b_3 & \dots \\ a_2b_0 & a_2b_1 & a_2b_2 & a_2b_3 & \dots \\ a_1b_0 & a_1b_1 & a_1b_2 & a_1b_3 & \dots \\ a_0b_0 & a_0b_1 & a_0b_2 & a_0b_3 & \dots \\ \end{array} $$ What do you notice about the diagonals?

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