Confused By Proof: Baby Rudin Chapter 3 Exercise 13

I found a proof on StackExchange for Exercise 13, Chapter 3 in Baby Rudin:

Problem:

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

Here is the proof somebody posted:

Proof:

I am confused how they got to the step marked in the red box.

Thank you.

$\displaystyle\sum_{k=0}^{n}\sum_{m=0}^{k}=\sum_{m=0}^{n}\sum_{k=m}^{n}$, then use $k'=k-m$, then $\displaystyle\sum_{k=m}^{n}|a_{m}b_{k-m}|=\sum_{k'=0}^{n-m}|a_{m}b_{k'}|$.

• So basically, $\displaystyle \sum_{m=0}^{n} \sum_{k=m}^{n}$ is doing an equivalent sum of $\displaystyle \sum_{k=0}^{n} \sum_{m=0}^{k}$, but backwards. And when we make the substitution $k^{\prime} = k - m$, we ask ourselves, "What is $k^{\prime}$ when $k = m$?" and "What is $k^{\prime}$ when $k = n$?" – Frederic Chopin Nov 13 '17 at 21:36

The boxed inequality has an index error. It should be

$$\sum^{n}_{k=0} \sum^{k}_{m=0} |a_m b_{k-m}| \leq \sum^{n}_{m=0} |a_m|\sum^{n-m}_{k=0} |b_k|$$

To see that the inequality holds, consider all the terms $|a_m b_{k-m}|$. For any given $m$, the indexes of $b$ range from $0$ to at most $n-m$. Also note that $|a_m b_{k-m}| = |a_m| |b_{k-m}|$.

• I have another question. When they write $\displaystyle \sum^{n}_{k=0} \sum^{k}_{m=0} |a_m b_{k-m}| \leq \sum^{n}_{m=0} |a_m|\sum^{n-m}_{k=0} |b_k|$, do they mean $\displaystyle \sum^{n}_{k=0} \sum^{k}_{m=0} |a_m b_{k-m}| \leq \left(\sum^{n}_{m=0} |a_m|\right)\left(\sum^{n-m}_{k=0} |b_k|\right)$, or do they mean $\displaystyle \sum^{n}_{k=0} \sum^{k}_{m=0} |a_m b_{k-m}| \leq \sum^{n}_{m=0}\left[|a_m|\sum^{n-m}_{k=0} |b_k|\right]$? – Frederic Chopin Nov 13 '17 at 21:39
• Walter Rudin obviously means the latter, since $m$ that is the index variable of the outer sum occurs in the bound of the inner sum. – Hans Hüttel Nov 13 '17 at 22:31
• But in that case, how can whoever writing this proof say that $\displaystyle \sum^{n}_{m=0}\left[|a_m|\sum^{n-m}_{k=0} |b_k|\right] \leq \left(\sum^{n}_{m=0} |a_m|\right)B$? – Frederic Chopin Nov 14 '17 at 1:34
• Remember that, since all summands are positive, we have that $\sum^{n-m}_{k=0} |b_k| \leq \sum^{n}_{k=0}$ for all $m \geq 0$. – Hans Hüttel Nov 14 '17 at 7:09
• Ahh, OK, thank you :) – Frederic Chopin Nov 14 '17 at 20:16