# Proof

We begin the proof with a number of definitions: We set

$\begin{array}{c} A:=\sum_{n=0}^\infty a_n,\quad A_n:=\sum_{k=0}^n a_k\\ B:=\sum_{n=0}^\infty b_n,\quad B_n:=\sum_{k=0}^n b_k\\ c_n:=\sum_{k=0}^n a_{n-k}b_k,\quad C_n:=\sum_{k=0}^n c_k\\ D_n:=\left(\sum_{k=0}^n a_k\right)\left(\sum_{k=0}^n b_k\right)=A_n\cdot B_n\\ D_n^*:=\left(\sum_{k=0}^n |a_k|\right)\left(\sum_{k=0}^n |b_k|\right). \end{array}$

The proof is in two steps:

### $$(1)$$ We will first show that $$\lim_{n\to\infty} C_n=A\cdot B$$.

By hypothesis, $$A$$ and $$B$$ are absolutely convergent series, which by definition means that $$\sum_{n=0}^\infty |a_n|$$ and $$\sum_{n=0}^\infty |b_n|$$ are convergent series. Due to the rule of the product of convergent sequences, it follows that the real sequences $$(D_n^*)_{n\in\mathbb N}$$ is a convergent sequence. This in turn means by definition that for a fixed $$\epsilon > 0$$ there is an index $$m\in\mathbb N$$ with $|D_n^*- D_m^*| < \epsilon$

for all $n\ge m$ and, in particular, for all $n > 2m.$ Note that if $$n > 2m$$, then1 $\begin{array}{rcl} D_n^* - D_m^*&=&\sum_{\substack{0\le i\le n \\ 0\le j\le n}} |a_i||b_j|-\sum_{\substack{0\le i\le m \\ 0\le j\le m}} |a_i||b_j|\\ &=&\sum_{i,j} |a_i||b_j|[0\le i\le n\wedge 0\le j\le n]-\sum_{i,j}|a_i||b_j|[0\le i\le m\wedge 0\le j\le m]\\ &=&\sum_{i,j} |a_i||b_j|[m < i \le n\wedge m < j\le n].\quad\quad ( * ) \end{array}$

We will use this as an interim result.

Now, observe that first: $D_n=\left(\sum_{k=0}^n a_k\right)\left(\sum_{k=0}^n b_k\right)=\sum_{\substack{0\le i\le n \\ 0\le j\le n}} a_ib_j=\sum_{i,j} a_ib_j[0\le i\le n\wedge 0\le j\le n]$ and that second $C_n=\sum_{k=0}^n c_k=\sum_{k=0}^n \sum_{j=0}^k a_{k-j}b_j=\sum_{k=0}^n \sum_{i+j=k} a_ib_j=\sum_{i+j\le n} a_ib_j=\sum_{i,j} a_ib_j[i+j\le n]$ and that third, according to the triangle inequality and the properties of the absolute value. $\begin{array}{rcl} |D_n - C_n|&=&\left|\sum_{i,j} a_ib_j[0\le i\le n\wedge 0\le j\le n] - \sum_{i,j} a_ib_j[i+j\le n]\right|\\ &=&\left|\sum_{i,j} a_ib_j[0\le i\le n\wedge 0\le j\le n\wedge i+j > n]\right|\\ &\le &\sum_{i,j} |a_i||b_j|[0\le i\le n\wedge 0\le j\le n\wedge i+j > n]\quad\quad ( * * ) \end{array}$ This is our second interim result.

Now, once again since $$A$$ and $$B$$ are absolutely convergent series, they are in particular convergent. Therefore, the real sequences of their partial sums $$(A_n)_{n\in\mathbb N}$$ and $$(B_n)_{n\in\mathbb N}$$ are convergent sequences with $$\lim A_n=A$$ and $$\lim B_n=B$$. Like in the reasoning above, due to the rule of the product of convergent sequences, we have $\lim_{n\rightarrow\infty} D_n=\lim_{n\rightarrow\infty} (A_n \cdot B_n)=\lim_{n\rightarrow\infty} A_n \cdot \lim_{n\rightarrow\infty} B_n=A\cdot B.$ Thus, it suffices to demonstrate that $\lim_{n\to\infty} (D_n - C_n)=0.$ But this will follow immediately, if we can show that for a given $$\epsilon > 0$$ and all $$n > 2m$$ the set of possible index pairs $$\{(i,j)\}$$ in the sum $$( * * )$$ is included in the set of index pairs $$\{(i,j)\}$$, which are possible in the sum $$( * )$$. If this is the case, the inequality $|D_n - C_n| \le |D_n^*- D_m^*| < \epsilon\text{ for all }n\ge 2m$ will be true. But the set of possible index pairs $$\{(i,j)\}$$ in the sum $$( * * )$$ is indeed included in the set of index pairs $$\{(i,j)\}$$, which are possible in the sum $$( * )$$, like demonstrated in the following figure This is a pure geometric argument, since for $$n > 2m$$ the hatched area is totally included in the blue area. Therefore, we have $$\lim_{n\to\infty} (D_n - C_n)=0,$$ and $$\lim_{n\to\infty} C_n=A\cdot B$$.

### $$(2)$$ It remains to be shown $$\sum_{n\to\infty} c_n$$ is absolutely convergent.

We have to show that $$\sum_{n\to\infty}|c_n|$$ is convergent. It has already been shown that the real sequence $$(D_n^*)_{n\in\mathbb N}$$ is a convergent sequence. Moreover, we have in the Iverson notation

$D_n^*=\sum_{i,j} |a_i||b_j|[0\le i\le n\wedge 0\le j\le n].\quad\quad ( \times )$

On the other hand, it follows from the triangle inequality that $\sum_{k=0}^n |c_k|=\sum_{k=0}^n \left|\sum_{j=0}^k a_{k-j}b_j\right|\le \sum_{k=0}^n \sum_{i+j=k} |a_i||b_j|=\sum_{i,j} |a_i||b_j|[i+j\le n]\quad\quad ( \times\times )$

Following a similar geometric argument as above, the set of of possible index pairs $$\{(i,j)\}$$ in the sum $$( \times\times )$$ is included in the set of index pairs $$\{(i,j)\}$$, which are possible in the sum $$( \times )$$ - see following figure Therefore, we have

$\sum_{k=0}^n |c_k|\le D_n^*$ and it follows from the majorant criterion that $$\sum_{n\to\infty}|c_n|$$ is convergent.

Github: #### Footnotes

1. In the following text, the Iverson notation for sum is used, which turns out to be very helpful in sum manipulation steps.