Cauchy product proof 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:

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.
 A: While @Siminore's question is illuminating as to the first part of the OP's question, it doesn't answer what I see as the OP's main question, of why the Cauchy product has that form, or why the finite sum he posted is incorrect. In fact, both answers have nothing to do with power series.
To use the graphic posted by @user365294 , we can see that the original form of the summation corresponds to summing row-by-row or column-by-column, whereas the cauchy form is equivalent to summing along the diagonals from the top left to the bottom right -- notice that these diagonals grow in size, which is why you have this asymmetry in the limits of the summation. 
As to why the finite sum the OP posted does not show this pattern, it is precisely because it is finite. Notice that if you are summing over a finite "square" in this lattice, the diagonals do not keep growing in the same way towards infinity, because at some point you get past the main diagonal, so they begin to shrink again. Therefore the correct formula for the finite sum over a square would be 
$$(\sum_{i=0}^{L} a_i)(\sum_{i=0}^{L} b_i)$$
$$=\sum_{i=0}^{L}\sum_{j=0}^{L} a_i*b_j$$
$$=\sum_{i=0}^{L}\sum_{j=0}^{i} a_j*b_{i-j}
+\sum_{i=1}^{L}\sum_{j=0}^{i} a_j*b_{L-i+j}$$
So basically you are missing a term in the correct expression for your finite example. If you plug this finite version of Cauchy's formula in to your example, it will evaluate to the correct answer.
A: 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?
A: The pattern is suggested by the multiplication of power series. If you consider the two power series
$$
\sum_{k=0}^\infty a_k x^k, \quad \sum_{k=0}^\infty b_k x^k,
$$
then their product is the power series $\sum_{k=0}^\infty c_k x^k$ such that
$$
c_k = \sum_{j=0}^k a_j b_{k-j}.
$$
