# Natural Deductions

Expository preaching, non-rigorous proofs

## Thursday, 22 April 2010

### Cauchy Product

We have two convergent series,

$A_n = \sum^n_{k=0} a_k$ & $B_n = \sum^n_{k=0} b_k$.

Theorem: Cauchy Product $C_\infty = A_\infty B_\infty = \sum_{k=0} \sum_{i=0..k} a_i b_{k-i}$.

Demonstration: The product is the sum of the sums anti-diagonals of this table:

$\begin{tabular}{ c | c | c | c | c } (a_0 b_0) & (a_1 b_0) & (a_2 b_0) & (a_3 b_0) & a_4 b_0 & \\ \hline (a_0 b_1) & (a_1 b_1) & (a_2 b_1) & a_3 b_1 & a_4 b_1 & \\ \hline (a_0 b_2) & (a_1 b_2) & a_2 b_2 & a_3 b_2 & a_4 b_2 & \cdots \\ \hline (a_0 b_3) & a_1 b_3 & a_2 b_3 & a_3 b_3 & a_4 b_3 & \\ \hline a_0 b_4 & a_1 b_4 & a_2 b_4 & a_3 b_4 & a_4 b_4 & \\ \hline & & \vdots & & \ddots & \\ \end{tabular}$

Bracketed is $C_3$, now $A_n B_n \leq C_{2n} \leq A_{2n} B_{2n}$ so $C_\infty$ converges. $\qed$

Cauchy Product - Planet Math
The theorem of Mertens about the Cauchy product of infinite series - Harold P. Boas