Cauchy product of two absolutely convergent series is absolutely convergent. (Rudin PMA Ch. 3 ex 13)

Given $\sum a_n$ and $\sum b_n$ we define the Cauchy product to be $\sum_{k = 0}^{n}a_kb_{n-k}$. I need to prove that if both $\sum a_n$ and $\sum b_n$ are absolutely convergent then so is the Cauchy product.

Unless I'm missing something the proof seems trivial to me. Theorem 3.50 in the book "baby Rudin" states that given two convergent series at least one of which is absolutely convergent the Cauchy product will converge to the product of the limits. So using this why not just say the following?

Since $\sum_{n=0}^\infty |a_n|$ is convergent it is also absolutely convergent. Therefore, by the theorem 3.50,

$$\sum_{n=0}^\infty \sum_{k=0}^{n}|a_kb_{n-k}| = \sum_{n=0}^\infty |a_n| \sum_{n=0}^\infty |b_n|$$

Proof: Since $\sum_n |a_n|$ and $\sum_n |b_n|$ converges, let $\sum_n |a_n| < M$ and $\sum_n |b_n| < N$

$$\sum_{n = 0}^m |c_n| = \sum_{n = 0}^m\left|\sum_{k = 0}^n a_kb_{n-k}\right|\le\sum_{n = 0}^m\sum_{k = 0}^n |a_kb_{n-k}| $$ $$= |a_0b_0|+(|a_0b_1|+|a_1b_0|)+\cdots+(|a_0b_m|+|a_1b_{m-1}|+\cdots + |a_mb_0|)$$ $$ = \sum_{n = 0}^m |a_n| \sum_{k = 0}^{m-n} |b_k|< \sum_{n = 0}^m |a_n|N<NM$$ $\sum_n |c_n|$ is bounded above and monotone (since its sum of non-negative terms), it converges.
Thus Cauchy product of two absolutely convergent series converges absolutely.

Proof: Let $\sum a_n$ and $\sum b_n$ be absolutely convergent with $\sum a_n=:A\in\mathbb{R}$ and $\sum b_n=:B\in\mathbb{R}$. We claim that $\sum c_n=\sum_n (\sum_{m=0}^{n} a_m b_{n-m})$ too is absolutely convergent and it converges to $AB$.

Set $\forall n: A_n:= \sum_{k=0}^{n} a_k, B_n:= \sum_{k=0}^{n} b_k, C_n:= \sum_{k=0}^{n} c_k,$ $\tilde{A}_n:= \sum_{k=0}^{n} |a_k|, \tilde{B}_n:= \sum_{k=0}^{n} |b_k|, \tilde{C}_n:= \sum_{k=0}^{n} |c_k|$, and consider the following $2n\times 2n$ matrix:

\begin{bmatrix} a_0b_0 & a_1b_0 & \cdots & a_nb_0 & \cdots & \cdots & a_{2n}b_{0}\\ a_0b_1 & a_1b_1 & \cdots & a_nb_1 & \cdots & \cdots & a_{2n}b_{1}\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\ a_0b_n & a_1b_n & \cdots & a_nb_n & \cdots & \cdots & a_{2n}b_{n}\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\ a_0b_{2n} & a_1b_{2n} & \cdots & a_nb_{2n} & \cdots & \cdots & a_{2n}b_{2n}\\ \end{bmatrix}

Then the upper left $n\times n$ submatrix consists of all the terms of $A_nB_n$, while "anti-diagonal"s are the terms of $c_n$ (so that $c_0$ is the sum of the terms on the first "anti-diagonal", $c_1$ is the sum of the terms on the second "anti-diagonal" and so on), and consequently $C_n$ is the sum of the first $n$ "anti-diagonal"s.

We have two cases: If $\{a_n\}_n,\{b_n\}_n\subseteq \mathbb{R}_{\geq0}$, then by the above interpretation of the matrix we have $C_n\leq A_nB_n\leq C_{2n}$. If, on the other hand, $\{a_n\}_n,\{b_n\}_n\subseteq \mathbb{R}$ are arbitrary sequences, then considering the matrix $(|a_ib_j|)_{i,j\leq n}$ we have $\tilde{C}_n\leq \tilde{A}_n\tilde{B}_n\leq \tilde{C}_{2n}$.

In the first case, $A_n\uparrow A$ and $B_n\uparrow B$. Then $\forall n: C_n\leq A_nB_n\leq C_{2n} \implies \limsup C_n\leq AB \leq \limsup C_{2n}$. Since $\{C_{2n}\}_n\subseteq \{C_{n}\}_n, \limsup C_{2n}\leq \limsup C_{n}$, and hence $\limsup C_n=AB$. As $\{C_n\}_n\uparrow$ too, it can have at most one (real) subsequential limit, viz. its superior limit. Thus $C_n\uparrow AB (\implies \tilde{C}_n= C_n\uparrow AB)$.

For the general case observe that $|A_nB_n-C_n|\leq \tilde{A}_n\tilde{B}_n-\tilde{C}_n$ (in the above matrix these correspond to the sums of the terms of the $n\times n$ matrix that are below the anti-diagonal). The right-hand side vanishes as $n\to\infty$ by the preceding argument (since we have series with nonnegative terms). Then $\lim |A_nB_n-C_n|=0 \implies \lim C_n =\lim A_nB_n= AB$. Q.E.D.

You did not write the Cauchy product correctly - where is the sum over $k$? When you fix this, you will need the triangle inequality and comparison test to finish the proof.

Yes, this is an easy exercise if you use Theorem 3.50. Arguably, it is a better exercise if you don't use Theorem 3.50 - a direct proof of the boundedness of partial sums is short and sweet.