Question

prove that the Cauchy product of two absolutely convergent series converges absolutely

Answer 1

$\frac{f(b) - f(a) }{g(b) - g(a)} = \frac{f(c)}{g(c)}$

Answer 2

Answer:

The statement shows $\sum\left|c_{n}\right| \leq s_{n}$ and since $s_{n}$ converges, $\sum c_{n}$ must also converge absolutely.

Step-by-step explanation:

The Cauchy product of an absolutely convergent series converges.

Let $\sum a_{n}$ and $\sum b_{n}$ be two absolutely convergent series. Then sequence

$s_{n}=\sum_{i=1}^{n} \sum_{j=1}^{n}\left|a_{i} b_{j}\right|$

converges. Suppose $\sum\left|a_{n}\right|=\alpha$ and $\sum\left|b_{n}\right|=\beta$ then

$s_{n}=\sum_{i=1}^{n}\left|a_{i}\right| \sum_{j=1}^{n}\left|b_{j}\right| \leq \alpha \beta$

and $\left\{s_{n}\right\}$ being a monotone sequence that is increasing and bounded by above must converge. Define

$c_{n}=\sum_{k=1}^{n} a_{k} b_{n-k}$

Then $\sum c_{n}$ converges absolutely. We have:

$\left|c_{n}\right|=\left|\sum_{k=1}^{n} a_{k} b_{n-k}\right| \leq \sum_{k=1}^{n}\left|a_{k} b_{n-k}\right|$

and

$\sum_{k=1}^{n}\left|c_{k}\right| \leq \sum_{k=1}^{n} \sum_{j=1}^{k}\left|a_{j} b_{k-j}\right| \leq \sum_{j=1}^{n} \sum_{k=1}^{n}\left|a_{j} b_{k}\right|$

This shows $\sum\left|c_{n}\right| \leq s_{n}$ and since $s_{n}$ converges, $\sum c_{n}$ must also converge absolutely.