# Exercise from (Baby) Rudin (Chapter 3, exercise 13): is this proof correct? Is it well-written?

The problem is the following:

Prove that the Cauchy product of two absolutely convergent series converges absolutely.

Here is my attempt:

Let $s_n=\sum^n_{k=0}a_k$ and $t_n=\sum^n_{k=0}b_k$ be two series absolutely converging to $A$ and $B$ respectively; then, let $c_n=\sum^n_{k=0}a_kb_{n-k}$ and $C_n=\sum^n_{k=0}c_k$. We have to prove that the series $D_n=\sum^n_{k=0}|c_k|$ converges.

Since $D_n$ is a series of non-negative terms, it will suffice to show that it is bounded. And, in fact, $$D_n=|a_0b_0|+|a_0b_1+a_1b_0|+...+|a_0b_n+...+a_nb_0|\leq(|a_0||b_0|)+(|a_0||b_1|+|a_1||b_0|)+...+(|a_0||b_n|+...+|a_n||b_0|).$$

The RHS of the inequality is the Cauchy product of the series $\sum^n_{k=0}|a_k|$ and $\sum^n_{k=0}|b_k|$, both of which converge by hypothesis. Then, since $\sum^n_{k=0}||a_k||= \sum^n_{k=0}|a_k|$ and $\sum^n_{k=0}||b_k||=\sum^n_{k=0}|b_k|$, they also converge absolutely. The Cauchy product of two absolutely convergent series converges and, therefore, is bounded. Hence, $D_n$ is also bounded and, thus, convergent.

It seems fine to me, but also a little too easy perhaps! Assuming that it is correct...is it written properly? Should I specify some steps a little more? Should I be more explicit or succinct? Would this be accepted at a university exam?

• It is helpful to mention which exercise this is. – Tim Raczkowski Aug 30 '15 at 22:05
• This seems fine to me, as long as you're allowed to use that the Cauchy product of two absolutely convergent sequence converges. I can't speak for the exams, but I'd accept it. – Patrick Stevens Aug 30 '15 at 22:06
• @TimRaczkowski I've edited the question! It's the exercise 13, chapter 3. – Nicol Aug 30 '15 at 22:08
• @PatrickStevens Yes, the theorem was included in the theory section before the problems. – Nicol Aug 30 '15 at 22:08

Since, $\sum |c_n|$ is a sequence of non-negative terms, it suffices to show that $\{C_n\}$ is bounded.
\begin{align} C_n & =\sum_{k=0}^n\left|\sum_{m=0}^k a_mb_{k-m}\right|\\ & \le\sum_{k=0}^n\sum_{m=0}^k|a_mb_{k-m}|\\ & =\sum_{m=0}^n|a_n|\sum_{k=0}^{n-m}|b_k|\\ &\le\left(\sum_{k=0}^n|a_k|\right)B\le AB. \end{align}
Here, $C_n$ is a partial sum and $A$ and $B$ are the limits of $\sum a_n$ and $b_n$ respectively.