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Given $\sum a_n$ and $\sum b_n$ we define the Cauchy product to be $\sum_{k = 0}^{n}a_kb_{n-k}$. I need to prove that if both $\sum a_n$ and $\sum b_n$ are absolutely convergent then so is the Cauchy product.

Unless I'm missing something the proof seems trivial to me. Theorem 3.50 in the book "baby Rudin" states that given two convergent series at least one of which is absolutely convergent the Cauchy product will converge to the product of the limits. So using this why not just say the following?

Since $\sum_{n=0}^\infty |a_n|$ is convergent it is also absolutely convergent. Therefore, by the theorem 3.50,

$$\sum_{n=0}^\infty \sum_{k=0}^{n}|a_kb_{n-k}| = \sum_{n=0}^\infty |a_n| \sum_{n=0}^\infty |b_n|$$

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  • $\begingroup$ Where did this random picture come from? $\endgroup$ – AAA Sep 25 '12 at 22:07
  • $\begingroup$ If you're talking about the infant, it is maybe somepicture you use in other pages. Gravatars might sync in openIDs, I guess. $\endgroup$ – Pedro Tamaroff Sep 25 '12 at 22:42
  • $\begingroup$ what you wrote is not cauchy product. $\endgroup$ – mez Mar 10 '13 at 17:44
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Proof: Since $\sum_n |a_n|$ and $\sum_n |b_n|$ converges, let $\sum_n |a_n| < M$ and $\sum_n |b_n| < N$

$$\sum_{n = 0}^m |c_n| = \sum_{n = 0}^m|\sum_{k = 0}^n a_kb_{n-k}|\le\sum_{n = 0}^m\sum_{k = 0}^n |a_kb_{n-k}| $$ $$= |a_0b_0|+(|a_0b_1|+|a_1b_0|)+\cdots+(|a_0b_m|+|a_1b_{m-1}|+\cdots + |a_mb_0|)$$ $$ = \sum_{n = 0}^m |a_n| \sum_{k = 0}^{m-n} |b_k|< \sum_{n = 0}^m |a_n|N<NM$$ $\sum_n |c_n|$ is bounded above and monotone (since its sum of non-negative terms), it converges.
Thus Cauchy product of two absolutely convergent series converges absolutely.

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Proof: Let $\sum a_n$ and $\sum b_n$ be absolutely convergent with $\sum a_n=:A\in\mathbb{R}$ and $\sum b_n=:B\in\mathbb{R}$. We claim that $\sum c_n=\sum_n (\sum_{m=0}^{n} a_m b_{n-m})$ too is absolutely convergent and it converges to $AB$.

Set $\forall n: A_n:= \sum_{k=0}^{n} a_k, B_n:= \sum_{k=0}^{n} b_k, C_n:= \sum_{k=0}^{n} c_k,$ $\tilde{A}_n:= \sum_{k=0}^{n} |a_k|, \tilde{B}_n:= \sum_{k=0}^{n} |b_k|, \tilde{C}_n:= \sum_{k=0}^{n} |c_k|$, and consider the following $2n\times 2n$ matrix:

\begin{bmatrix} a_0b_0 & a_1b_0 & \cdots & a_nb_0 & \cdots & \cdots & a_{2n}b_{0}\\ a_0b_1 & a_1b_1 & \cdots & a_nb_1 & \cdots & \cdots & a_{2n}b_{1}\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\ a_0b_n & a_1b_n & \cdots & a_nb_n & \cdots & \cdots & a_{2n}b_{n}\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\ a_0b_{2n} & a_1b_{2n} & \cdots & a_nb_{2n} & \cdots & \cdots & a_{2n}b_{2n}\\ \end{bmatrix}

Then the upper left $n\times n$ submatrix consists of all the terms of $A_nB_n$, while "anti-diagonal"s are the terms of $c_n$ (so that $c_0$ is the sum of the terms on the first "anti-diagonal", $c_1$ is the sum of the terms on the second "anti-diagonal" and so on), and consequently $C_n$ is the sum of the first $n$ "anti-diagonal"s.

We have two cases: If $\{a_n\}_n,\{b_n\}_n\subseteq \mathbb{R}_{\geq0}$, then by the above interpretation of the matrix we have $C_n\leq A_nB_n\leq C_{2n}$. If, on the other hand, $\{a_n\}_n,\{b_n\}_n\subseteq \mathbb{R}$ are arbitrary sequences, then considering the matrix $(|a_ib_j|)_{i,j\leq n}$ we have $\tilde{C}_n\leq \tilde{A}_n\tilde{B}_n\leq \tilde{C}_{2n}$.

In the first case, $A_n\uparrow A$ and $B_n\uparrow B$. Then $\forall n: C_n\leq A_nB_n\leq C_{2n} \implies \limsup C_n\leq AB \leq \limsup C_{2n}$. Since $\{C_{2n}\}_n\subseteq \{C_{n}\}_n, \limsup C_{2n}\leq \limsup C_{n}$, and hence $\limsup C_n=AB$. As $\{C_n\}_n\uparrow$ too, it can have at most one (real) subsequential limit, viz. its superior limit. Thus $C_n\uparrow AB (\implies \tilde{C}_n= C_n\uparrow AB)$.

For the general case observe that $|A_nB_n-C_n|\leq \tilde{A}_n\tilde{B}_n-\tilde{C}_n$ (in the above matrix these correspond to the sums of the terms of the $n\times n$ matrix that are below the anti-diagonal). The right-hand side vanishes as $n\to\infty$ by the preceding argument (since we have series with nonnegative terms). Then $\lim |A_nB_n-C_n|=0 \implies \lim C_n =\lim A_nB_n= AB$. Q.E.D.

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  • $\begingroup$ Wow! Thanks for the 'visual' aid in the proof, by providing matrix. this was amazing way to see Cauchy product. $\endgroup$ – Silent Mar 2 '18 at 12:00
  • $\begingroup$ Hey, may you please explain why $A_n \cdot B_n \le C_{2n}$ ? (for the non negative part) cause for example, if we assume $a_0 < a_1 < a_2 <...$ and $b_0 < b_1 < b_2 < ... $ in case $n=1$ , $A_1 \cdot B_1 = a_0 b_0 + a_0 b_1 + a_1 b_0 +a_1 b_2 $ and $C_2 = a_0 b_2 + a_1 b_1 +a_2 b_0$ and I don't see why $A_1 B_1 \le C_2$ in that case. Thanks! $\endgroup$ – dan Dec 28 '18 at 7:04
  • $\begingroup$ For the case you are citing we are not assuming that the terms are increasing, they are just nonnegative. The last summand of $A_1B_1$ is $a_1b_1$. Finally $C_2=c_0+c_1+c_2$. $\endgroup$ – Alp Uzman Dec 30 '18 at 14:05
  • $\begingroup$ Alternatively you can look at the matrix I wrote above. When all terms are nonnegative, since $C_{2n}$ is the sum of all terms of the $2n\times 2n$ matrix above, and since $A_nB_n$ is the sum of all terms of the top left $n\times n$ part of it the claim follows. $\endgroup$ – Alp Uzman Dec 30 '18 at 14:05
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You did not write the Cauchy product correctly - where is the sum over $k$? When you fix this, you will need the triangle inequality and comparison test to finish the proof.

Yes, this is an easy exercise if you use Theorem 3.50. Arguably, it is a better exercise if you don't use Theorem 3.50 - a direct proof of the boundedness of partial sums is short and sweet.

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