# Question about Proof of Merten's theorem (Cauchy-Product formula)

I have a question about the proof on the german wikepedia page:

The proof is stated as follow:

Let $$A= \sum_{k=0}^{\infty}a_k$$ and $$B=\sum_{k=0}^{\infty}b_k$$, if at least one of them is absolutely convergent, then their Cauchy-Product converges to $$AB$$.

Definition of the Cauchy-Product: $$C=\sum_{k=0}^{\infty}c_k,c_k=\sum_{j=0}^{k}a_jb_{k-j}$$

Without loss of generality let A be the absolutely convergent series and $$S_n=\sum_{k=0}^{n}c_k$$

1: $$AB=(A-A_n)B+\sum_{k=0}^{n}a_kB$$

2: $$S_n=\sum_{k=0}^{n}a_kB_ {n-k}$$

1-2=$$AB-S_n=(A-A_n)B+\sum_{k=0}^{n}a_k(B-B_{n-k})$$

$$(A-A_n)B \rightarrow 0$$ and with $$N:=[\frac{n}{2}]$$ the other series can be splitted into two parts with:

$$\sum_{k=0}^{N}a_k(B-B_{n-k})+\sum_{k=N+1}^{n}a_k(B-B_{n-k})$$

Then

$$|\sum_{k=0}^{N}a_k(B-B_{n-k})|\leq \sum_{k=0}^{N}|a_k(B-B_{n-k})|=\sum_{k=0}^{N}|a_k||(B-B_{n-k})|\leq\max\limits_{N \leq k \leq n}|B-B_k|\sum_{k=0}^{N}|a_k|\rightarrow 0$$

Because the last expression of the above inequalities is a product with a zero-convergent sequence with a bounded sequence. Because the zero-convergent sequence $$(B-B_k)$$ is bounded there is a $$C > 0$$ with $$|B-B_k|

Hence

$$|\sum_{k=N+1}{n}a_k(B-B_{n-k})|\leq \sum_{k=N+1}{n}|a_k||(B-B_{n-k})|\leq C\sum_{k=N+1}{n}|a_k|\rightarrow 0 \square$$

I don't understand why the sum is splitted in two parts, can also somebody tell me what's with the $$max$$ estimate.

Thank you for your time, I would appreciate your help very much.

The reason for the splitting is that when $$k$$ is 'large" one gets that the $$a_k$$ are 'small', while when $$k$$ is 'small' then $$n-k$$ is 'large' and $$B - B_{n-k}$$ is 'small.'
Thus, depending on whether $$k$$ is 'large' or $$k$$ is 'small' different types of arguments work. However, it's not needed to split exactly in the middle, but it's a natural choice.
As for the max. The task is to estimate a sum of the form $$\sum_{k} |f_k| \ |g_k|$$ where one knows that $$\sum_{k} |f_k|$$ tends to $$0$$. To get rid of the $$|g_k|$$ one takes a $$G$$ that bounds $$|g_k| \le G$$ for each $$k$$, and estimates
$$\sum_{k} |f_k| \ |g_k| \le \sum_{k} |f_k| G = G \sum_{k} |f_k|$$