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