# 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 '16 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 $u_n=u_{n+1}=0$. – Brian M. Scott Oct 30 '16 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. – Saaqib Mahmood Oct 31 '16 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?