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.

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