## Rudin- Ch. 3 #13

Problem Statement: Given $\sum a_n$ and $\sum b_n$ define the product to be $\sum c_n$ where $c_n=\sum\limits_{k=0}^{n}a_kb_{n-k}$. Suppose that $\sum a_n$ converges to $A$ absolutely and $\sum b_n$ converges to $B$ absolutely. Prove that $\sum c_n$ converges to a value $C$ absolutely.

Proof: Since $\sum a_n$ and $\sum b_n$ converge absolutely we know that their product converges to $AB$ (this is by Rudin Theorem 3.50 which requires only one of the sums converge absolutely in order to guarantee that their product will converge to $AB$. This theorem does not guarantee that the product converges absolutely.).

Define $C_n=\sum\limits_{k=0}^{n}|c_k|$ where $c_k$ is as defined previously. Also define $\beta_n=B_n-B$ where $B_n=\sum\limits_{k=0}^{n}b_k$. Similarly let $A_n=\sum\limits_{k=0}^{n}a_k$. Then it follows that :

$C_n=\sum\limits_{k=0}^{n}\left|c_n\right|$

$=\left|a_0b_0+\left(a_0b_1+a_1b_0\right)+\dots+\left(a_0b_n+\dots+a_nb_0\right)\right|$

$=\left|a_0(b_0+\dots+b_n)+a_1(b_0+\dots+b_{n-1})+\dots\right|$

$\left|a_0B_n+a_1B_{n-1}+\dots+a_nB_0\right|$

$=\left|a_0(\beta_n+B)+a_1(\beta_{n-1}+B)+\dots+a_n(\beta_0+B)\right|$

$=\left|(a_0+\dots+a_n)B+a_0\beta_n+\dots+a_n\beta_0\right|$

$\leq\left|A_nB\right|+ \left|a_0\beta_n+\dots+a_n\beta_0 \right|$

Recall that we are trying to show that the sequence of $C_n$ converges to $AB$. So let $\gamma_n=\left|a_0\beta_n+\dots+a_n\beta_0\right|$. Since we know that $A_n\rightarrow A$ it follows that the limit of the above equation will tend to $AB$ iff $\gamma_n\rightarrow 0$. Let us define $A^{*}=\sum\limits_{n=0}^{\infty}\left|a_n\right|$ (we know that $A^{*}$ exists since we know the series converges absolutely).

First note that  as $n\rightarrow\infty$ it follows that $\beta_n\rightarrow 0$ since $B_n\rightarrow B$. So let $\varepsilon >0$. Then there is a $N\in\mathbb{N}$ such that for $n> N$ it follows that $\left|B_n-B\right|<\dfrac{\varepsilon}{A^{*}}$. But by our definition of $\beta_n$ this also means that for $n>N$ it follows that $\left|\beta_n\right|<\dfrac{\varepsilon}{A^{*}}$.

Now consider $n> N$:

$\left|\gamma_n\right|=\left|a_0\beta_n+\dots+a_n\beta_0\right|$

$\leq \left|a_0\beta_n+\dots+a_{n-N-1}\beta_{N+1}\right|+\left|a_{n-N}\beta_N+\dots+a_n\beta_0\right|$

$\leq\left|a_0\beta_n+\dots+a_{n=N-1}\beta_{N+1}\right|+\left|a_{n-N}\beta_N\right|+\dots+\left|a_n\beta_0\right|$

$<\left|a_0\beta_n+\dots+a_{n-N-1}\beta_{N+1}\right|+\left(\sum\limits_{k=0}^{n}\left|a_k\right|\right)\dfrac{\varepsilon}{A^{*}}$

$\leq\left|a_0\beta_n+\dots+a_{n-N-1}\beta_{N+1}\right|+A^{*}\dfrac{\varepsilon}{A^{*}}$

$=\left|a_0\beta_n+\dots+a_{n-N-1}\beta_{N+1}\right|+\varepsilon$

Now let $\delta=\left|a_0\beta_n+\dots+a_{n-N-1}\beta_{N+1}\right|$. Since $N$ is fixed the above inequality holds as we take the limit as $n\rightarrow\infty$. But we know that $a_n\rightarrow 0$ since the series converges and so as $n\rightarrow\infty$ it follows that $\delta\rightarrow 0$. This implies that $\left|\gamma_n\right|< \varepsilon$, but since $\varepsilon$ may be made arbitrarily small, it follows that $\left|\gamma_n\right|\rightarrow 0$.

Thus, we may conclude that $C_n\rightarrow AB$ and so $\sum \left|c_n\right|$ converges, thus, $\sum c_n$ converges absolutely.

$\Box$

Reflection: The key to this proof is that both original series are absolutely convergent. The necessity of this isn’t immediately clear, but once you try to show that $\gamma_n\rightarrow 0$ you see that you need to know that you can bound the absolute values of the terms from the series.