# Prob. 13, Chap. 3 in Baby Rudin: The Cauchy product of two absolutely convergent series converges absolutely

Here's Prob. 13, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

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

My effort:

Let $\sum a_n$, $\sum b_n$ be two absolutely convergent series of complex numbers, and let $c_n = a_0 b_n + a_1 b_{n-1} + \cdots + a_n b_0 = \sum_{j = 0}^n a_j b_{n-j}$ for $n = 0, 1, 2, 3, \ldots$.

Let $s = \sum_{n =0}^\infty \left\vert a_n \right\vert$ and $t = \sum_{n =0}^\infty \left\vert b_n \right\vert$. We show that the series $\sum \left\vert c_n \right\vert$ converges; that is, we show that $\sum_{n=0}^\infty \left\vert c_n \right\vert < +\infty$.

Now let's take $s_n = \sum_{k = 0}^n \left\vert a_k \right\vert$, $t_n = \sum_{k = 0}^n \left\vert b_k \right\vert$, and $u_n = \sum_{k = 0}^n \left\vert c_k \right\vert$. Then $\lim_{n \to \infty} s_n = s$ and $\lim_{n \to \infty} t_n = t$; moreover, the sequences $\left\{ s_n \right\}$ and $\left\{ t_n \right\}$ are monotonically increasing sequences of real numbers which are bounded above. So $$s = \sup \left\{ \ s_n \ \colon \ n \in \mathbb{N} \ \right\} \ \ \ \mbox{ and } \ \ \ t = \sup \left\{ \ t_n \ \colon \ n \in \mathbb{N} \ \right\}.$$

Thus, for each $n \in \mathbb{N}$, we have $u_n \leq u_{n+1}$ and \begin{align} \left\vert u_n \right\vert &= u_n = \left\vert c_0 \right\vert + \left\vert c_1 \right\vert + \left\vert c_2 \right\vert \cdots + \left\vert c_n \right\vert \\ &\leq \left\vert a_0\right\vert \left\vert b_0 \right\vert + \left( \left\vert a_0 \right\vert \left\vert b_1 \right\vert + \left\vert a_1 \right\vert \left\vert b_0 \right\vert \right) + \left( \left\vert a_0 \right\vert \left\vert b_2 \right\vert + \left\vert a_1 \right\vert \left\vert b_1 \right\vert + \left\vert a_2 \right\vert \left\vert b_0 \right\vert \right) + \cdots + \left( \left\vert a_0 \right\vert \left\vert b_n \right\vert + \left\vert a_1 \right\vert \left\vert b_{n-1} \right\vert + \left\vert a_2 \right\vert \left\vert b_{n-2} \right\vert + \cdots + \left\vert a_n \right\vert \left\vert b_0 \right\vert \right) \\ &= \left\vert a_0 \right\vert \left( \sum_{k=0}^n \left\vert b_k \right\vert \right) + \left\vert a_1 \right\vert \left( \sum_{k=0}^{n-1} \left\vert b_k \right\vert \right) + \left\vert a_2 \right\vert \left( \sum_{k=0}^{n-2} \left\vert b_k \right\vert \right) + \cdots + \left\vert a_{n-1} \right\vert \left( \sum_{k=0}^1 \left\vert b_k \right\vert \right) + \left\vert a_n \right\vert \left( \left\vert b_0 \right\vert \right) \\ &\leq \left( \sum_{k=0}^n \left\vert a_k \right\vert \right) \left( \sum_{k=0}^n \left\vert b_k \right\vert \right) \\ &= s_n t_n \\ &\leq st. \end{align} Thus, $\left\{ u_n \right\}$ is a monotonically increasing sequence of real numbers with ic bounded above. Hence this sequence converges.

Is this proof correct?

• It's also true if one series converges conditionally and the other converges absolutely.
– zhw.
Oct 30, 2016 at 17:14
• There’s just one very small error: for each $n\in\Bbb N$ we have $u_n\le u_{n+1}$, not $u_n<u_{n+1}$, since it’s possible that $u_n=u_{n+1}=0$. Oct 30, 2016 at 19:18
• @BrianM.Scott how wonderful it feels to hear from you, my dear sir!! Yes, you're right. I've just editted my post to incorporate the correction you've suggested. Oct 31, 2016 at 11:34

## 1 Answer

Suppose $\left\{a_n\right\}$, $\left\{b_n\right\}$, for $n = 0, 1, 2, 3, \ldots$, are sequences of complex numbers for which the series $\sum \left\vert a_n \right\vert$, $\sum \left\vert b_n \right\vert$ both converge. Let $c_n = \sum_{k=0}^n a_k b_{n-k}$, for $n = 0, 1, 2, 3, \ldots$. Then we have to show that the series $\sum \left\vert c_n \right\vert$ converges too.

Let $\alpha_n = \sum_{k=0}^n \left\vert a_k \right\vert$, let $\beta_n = \sum_{k=0}^n \left\vert b_k \right\vert$, and let $\gamma_n = \sum_{k=0}^n \left\vert c_k \right\vert$ for $n = 0, 1, 2, 3, \ldots$. Let $\alpha = \lim_{n \to \infty} \alpha_n$ and $\beta = \lim_{n \to \infty} \beta_n$. Then $\alpha$ and $\beta$ are both non-negative real numbers. Now we show that $\lim_{n\to\infty} \gamma_n$ exists and is $\leq \alpha \beta$.

We note that $$\gamma_0 = \left\vert c_0 \right\vert = \left\vert a_0 \right\vert \left\vert b_0 \right\vert = \alpha_0 \beta_0,$$ $$\gamma_1 = \left\vert c_0 \right\vert + \left\vert c_1 \right\vert \leq \left\vert a_0 \right\vert \left\vert b_0 \right\vert + \left( \left\vert a_0 \right\vert \left\vert b_1 \right\vert + \left\vert a_1 \right\vert \left\vert b_0 \right\vert \right) = \left\vert a_0 \right\vert \beta_1 + \left\vert a_1 \right\vert \beta_0 \leq \alpha_1 \beta_1,$$ \begin{align*} \gamma_2 & = \left\vert c_0 \right\vert + \left\vert c_1 \right\vert + \left\vert c_2 \right\vert \\ & \leq \left\vert a_0 \right\vert \left\vert b_0 \right\vert + \left( \left\vert a_0 \right\vert \left\vert b_1 \right\vert + \left\vert a_1 \right\vert \left\vert b_0 \right\vert \right) + \left( \left\vert a_0 \right\vert \left\vert b_2 \right\vert + \left\vert a_1 \right\vert \left\vert b_1 \right\vert + \left\vert a_2 \right\vert \left\vert b_0 \right\vert \right) \\ & = \left\vert a_0 \right\vert \beta_2 + \left\vert a_1 \right\vert \beta_1 + \left\vert a_2 \right\vert \beta_0 \\ & \leq \alpha_2 \beta_2, \end{align*}

and for $n = 3, 4, 5, \ldots$, we have \begin{align*} \gamma_n & = \left\vert c_0 \right\vert + \left\vert c_1 \right\vert + \cdots + \left\vert c_n \right\vert \\ &\leq \left\vert a_0 \right\vert \left\vert b_0 \right\vert + \left( \left\vert a_0 \right\vert \left\vert b_1 \right\vert + \left\vert a_1 \right\vert \left\vert b_0 \right\vert \right) + \cdots + \left( \left\vert a_0 \right\vert \left\vert b_n \right\vert + \left\vert a_1 \right\vert \left\vert b_{n-1} \right\vert + \cdots + \left\vert a_n \right\vert \left\vert b_0 \right\vert \right) \\ &= \left\vert a_0 \right\vert \beta_n + \left\vert a_1 \right\vert \beta_{n-1} + \cdots + \left\vert a_n \right\vert \beta_0 \\ &\leq \alpha_n \beta_n. \end{align*} Thus we have shown that $$\gamma_n \leq \alpha_n \beta_n$$ for $n = 0, 1, 2, 3, \ldots$.

Now as $\left\{ \alpha_n \right\}$, $\left\{ \beta_n \right\}$, for $n = 0, 1, 2, 3, \ldots$, are monotonically increasing sequences of real numbers, so we can conclude that $$\alpha = \sup \left\{ \ \alpha_n \ \colon \ n = 0, 1, 2, 3, \ldots \ \right\}, \ \ \beta = \sup \left\{ \ \beta_n \ \colon \ n = 0, 1, 2, 3, \ldots \ \right\}.$$ In particular, we have $0 \leq \alpha_n \leq \alpha$ and $0 \leq \beta_n \leq \beta$; so we also have $\gamma_n \leq \alpha \beta$, for $n = 0, 1, 2, 3, \ldots$.

Moreover, the sequence $\left\{ \gamma_n \right\}$, for $n = 0, 1, 2, 3, \ldots$, is a monotonically increasing sequence. Therefore this sequence converges and we also have $$\lim_{n \to \infty} \gamma_n \leq \alpha \beta.$$

Have I managed to get the proof right this time? If so, is there any way I can improve the presentation of it? If not, then where have I erred?