Product of Series

**slevvio** · November 20th 2009, 03:34 AM

Let $\sum_{n=0}^{\infty} a_n$ and $\sum_{n=0}^{\infty} b_n$ be absolutely convergent (complex) series with sums A and B respectively. For each n, define $c_n = \sum_{m=0}^{n} a_m b_{n-m}$ .

1. Show that $\sum_{n=0}^{\infty} c_n$ is absolutely convergent. [Hint: Follow the same basic plan as used in Prop 5.2 (a) ]

Now the proposition says : Suppose $\sum_{n=0}^{\infty} a_n$ is an absolutely convergent series (in $\mathbb{C}$ ) which has sum S. Then any rearrangement is also absolutely convergent and has sum S.

I am stumped I don't even have a beginning of an idea what to do

. There are even parts after this question too which make even less sense to me but I think tackling this bit is the first step. any help would be appreciated

**Opalg** · November 20th 2009, 04:38 AM

Originally Posted by slevvio

Let $\sum_{n=0}^{\infty} a_n$ and $\sum_{n=0}^{\infty} b_n$ be absolutely convergent (complex) series with sums A and B respectively. For each n, define $c_n = \sum_{m=0}^{n} a_m b_{n-m}$ .

1. Show that $\sum_{n=0}^{\infty} c_n$ is absolutely convergent. [Hint: Follow the same basic plan as used in Prop 5.2 (a) ]

Now the proposition says : Suppose $\sum_{n=0}^{\infty} a_n$ is an absolutely convergent series (in $\mathbb{C}$ ) which has sum S. Then any rearrangement is also absolutely convergent and has sum S.

I am stumped I don't even have a beginning of an idea what to do

. There are even parts after this question too which make even less sense to me but I think tackling this bit is the first step. any help would be appreciated

Write an array in which the element in the (i,j)-position is $|a_ib_j|$ . You will see that $c_n$ is the sum of the elements in the n'th northeast-southwest diagonal of this array:

$\begin{array}{ccccc}&\;\;c_0&\;\;c_1&\;\;c_2\\ &\swarrow &\swarrow &\swarrow\vspace{2ex}\\ |a_0b_0|&|a_0b_1|&|a_0b_2|&\ldots \\ |a_1b_0|&|a_1b_1|&|a_1b_2|&\ldots \\ |a_2b_0|&|a_2b_1|&|a_2b_2|&\ldots \end{array}$

The sum of all the elements in the first n+1 rows and columns is $S_n \stackrel{\text{d{ef}}}{=}\sum_{i=0}^n|a_i|\sum_{j =0}^n|b_j|$ . It should then be obvious from the above array that $S_{\lfloor n/2\rfloor}\leqslant \sum_0^nc_n\leqslant S_n$ . From that, you can deduce that $\textstyle\sum c_n$ is absolutely convergent.

**slevvio** · November 23rd 2009, 02:17 AM

OK thank you very much for the assistance I think I have constructed a proof for that part.

However I am struggling with the third part of this question:

2. Explain why $\sum_{r,s =0}^{n} a_r b_s \rightarrow AB$ [done]

3. Now show that $\sum_{n=0}^{\infty} c_n = AB$ .

HINT: [You might want to do this by estimating $| \sum_{r=0}^{2n} c_r -\sum_{r,s=0}^{n} a_r b_s |$ . Here you will have to use that $\sum_{n=0}^{\infty} a_n$ and $\sum_{n=0}^{\infty} b_n$ are absolutely convergent.]

I dont understand where the 2n is coming from or how to even use it. I think I am trying to prove the statement

$\forall \epsilon > 0 , \exists N \in \mathbb{N}$ such that $\forall n \ge N, | \sum_{r=0}^{n} c_r - AB | < \epsilon$ .

And I used the triangle inequality to get

$| \sum_{r=0}^{2n} c_r -\sum_{r,s=0}^{n} a_r b_s | \le |\sum_{r=0}^{2n} c_r | + \sum_{r=0}^{n} |a_r| \sum_{s=0}^{n} |b_s|$
Any help would be appreciated here this is quite frustrating

**Opalg** · November 23rd 2009, 08:38 AM

Originally Posted by slevvio

2. Explain why $\sum_{r,s =0}^{n} a_r b_s \rightarrow AB$ [done]

3. Now show that $\sum_{n=0}^{\infty} c_n = AB$ .

HINT: [You might want to do this by estimating $| \sum_{r=0}^{2n} c_r -\sum_{r,s=0}^{n} a_r b_s |$ . Here you will have to use that $\sum_{n=0}^{\infty} a_n$ and $\sum_{n=0}^{\infty} b_n$ are absolutely convergent.]

I dont understand where the 2n is coming from or how to even use it. I think I am trying to prove the statement

$\forall \epsilon > 0 , \exists N \in \mathbb{N}$ such that $\forall n \ge N, | \sum_{r=0}^{n} c_r - AB | < \epsilon$ .

Think about the array that I described in my previous comment. In that array, $\sum_{r,s =0}^{n} a_r b_s$ corresponds to the sum of the elements in the square region consisting of the first n+1 rows and columns, and $\sum_{r=0}^{2n} c_r$ is the sum of all the elements in the triangular region on and above the line r+s=n. That triangular region includes all the elements in the square region, plus elements in two smaller triangular regions. What you need to do is to show that the sum of the absolute values of the elements in those two smaller triangular regions becomes small as n gets large. This will show that $\sum_{r=0}^{2n} c_r$ is approximately equal to $\sum_{r,s =0}^{n} a_r b_s$ , which you have already shown converges to AB.

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