# Analysis of an absolutely convergence of series

• JD_PM
In summary: You proved convergence (not absolute convergence), but the proof is easy to fix (just leave something small out and it will work). For the rest, the idea of your proof is certainly correct, but I don't like how you wrote it down.
JD_PM

## Homework Statement

- Given a bounded sequence ##(y_n)_n## in ##\mathbb{C}##. Show that for every sequence ##(x_n)_n## in ##\mathbb{C}## for which the series ##\sum_n x_n## converges absolutely, that also the series ##\sum_n \left(x_ny_n\right)## converges absolutely.

- Suppose ##(y_n)_n## is a sequence in ##\mathbb{C}## with the following property: for each sequence ##(x_n)_n## in ##\mathbb{C}## for which the series ##\sum_n x_n## converges absolutely, also the series ##\sum_n \left(x_ny_n\right)## converges absolutely. Can you then conclude that ##(y_n)_n## is a bounded sequence? Explain!

## The Attempt at a Solution

I know the following theorem:

But I still do not know how to start with this problem. May you give me a hint?

I am not acquainted with mathematical proofs yet.

Thanks

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I will only answer problem 1. Homework policy asks for one problem at a time.

You have to prove that ##\sum_{n=1}^\infty x_ny_n## converges absolutely. I.e., ##\sum_{n=1}^\infty |x_n y_n|## must converge.

HINT: Use the comparison test on ##|x_ny_n|## using the fact that ##(y_n)## is bounded.

JD_PM
Math_QED said:
I will only answer problem 1. Homework policy asks for one problem at a time.

You have to prove that ##\sum_{n=1}^\infty x_ny_n## converges absolutely. I.e., ##\sum_{n=1}^\infty |x_n y_n|## must converge.

HINT: Use the comparison test on ##|x_ny_n|## using the fact that ##(y_n)## is bounded.

OK, this is what I would do: show that the partial sums of this series form a Cauchy sequence:

Given ##\epsilon > 0## there exists ##N## such that ##| \sum_{k = m + 1}^n x_k y_k | < \epsilon## if ##n > m \ge N##

JD_PM said:
OK, this is what I would do: show that the partial sums of this series form a Cauchy sequence:

Given ##\epsilon > 0## there exists ##N## such that ##| \sum_{k = m + 1}^n x_k y_k | < \epsilon## if ##n > m \ge N##

That's not what my hint suggests, but would work as well. Can you provide a full proof?

Math_QED said:
That's not what my hint suggests, but would work as well. Can you provide a full proof?
OK so this is what I got:

We are told that the series ##\sum_n x_n## converges absolutely, which means that:

$$\forall \varepsilon> 0, \exists N\in\mathbb{C} :\sum_{n=m}^{t} |x_n|<\varepsilon, \forall t>m> N$$

We also know this property of the absolute values:

$$\left|\sum x_n y_n\right|\le\sum|x_n y_n|$$

##(y_n)_n## is bounded, which means ##|y_n|\le B##; more explicitely:

$$\left|\sum\limits_{n}x_n y_n\right|\leq \sum\limits_{n}|x_n y_n| = \sum\limits_n |x_n||y_n|\leq \sup\limits_{k}|y_k|\sum\limits_{n}|x_n|< +\infty.$$

What leads to prove that the series ##\sum_n \left(x_ny_n\right)## converges absolutely:

$$\left|\sum x_n y_n\right|\le\sum|x_n y_n|\le B\sum|x_n| < \epsilon$$

JD_PM said:
OK so this is what I got:

We are told that the series ##\sum_n x_n## converges absolutely, which means that:

$$\forall \varepsilon> 0, \exists N\in\mathbb{C} :\sum_{n=m}^{t} |x_n|<\varepsilon, \forall t>m> N$$

We also know this property of the absolute values:

$$\left|\sum x_n y_n\right|\le\sum|x_n y_n|$$

##(y_n)_n## is bounded, which means ##|y_n|\le B##; more explicitely:

$$\left|\sum\limits_{n}x_n y_n\right|\leq \sum\limits_{n}|x_n y_n| = \sum\limits_n |x_n||y_n|\leq \sup\limits_{k}|y_k|\sum\limits_{n}|x_n|< +\infty.$$

What leads to prove that the series ##\sum_n \left(x_ny_n\right)## converges absolutely:

$$\left|\sum x_n y_n\right|\le\sum|x_n y_n|\le B\sum|x_n| < \epsilon$$

Note I am not using the comparison test on ##|x_ny_n|##. Could you show how to prove this using it? (Of course, if you consider what I provided is correct)

JD_PM said:
Note I am not using the comparison test on ##|x_ny_n|##. Could you show how to prove this using it? (Of course, if you consider what I provided is correct)

You proved convergence (not absolute convergence), but the proof is easy to fix (just leave something small out and it will work).

For the rest, the idea of your proof is certainly correct, but I don't like how you wrote it down.

Your proof consists of a bunch of nested quantors. That's not how such proofs are written down. Take a look at Rudin's proof of the theorem you wrote down. That's how mathematics should be written down (well, maybe with some more details). He fixes an ##\epsilon >0## at the beginning of the proof. For the rest of the proof, this epsilon remains fixed. He then obtains an element ##N## as in the definition of convergence and shows that it works.

Math_QED said:
You proved convergence (not absolute convergence).

Didn't I?. The series ##\sum_n \left(x_ny_n\right)## is said to converge absolutely if the series ##\left|\sum\limits_{n}x_n y_n\right|## converges. Proof in Rudin's book:

So my point is that as I proved that the series ##\left|\sum\limits_{n}x_n y_n\right|## converges:

$$\left|\sum x_n y_n\right|\le\sum|x_n y_n|\le B\sum|x_n| < \epsilon$$

Then we can assert that ##\sum_n \left(x_ny_n\right)## converges absolutely.

What am I missing here then?

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Math_QED said:
I don't like how you wrote it down.

Your proof consists of a bunch of nested quantors. That's not how such proofs are written down. Take a look at Rudin's proof of the theorem you wrote down. That's how mathematics should be written down (well, maybe with some more details). He fixes an ##\epsilon >0## at the beginning of the proof. For the rest of the proof, this epsilon remains fixed. He then obtains an element ##N## as in the definition of convergence and shows that it works.

OK so I tried to make an analogy of the problem I proposed and Rudin's 3.42 Theorem:

Choose ##B## such that ##|Y_n|\le B## for all ##n## (where ##Y_n## is the partial sum of ##\sum y_n##). Given ##\epsilon > 0##, there is an integer N such that ##y_n\le \frac{\epsilon}{2B}## for ##N \le m \le n##.

Then, Rudin's 3.42 Theorem develops from here.

Before carrying on I need to understand why:

$$y_n\le \frac{\epsilon}{2B}$$

Particularly the right hand side; why is ##\epsilon## divided by ##2B## ?

## 1. What is the definition of absolute convergence of a series?

A series is said to be absolutely convergent if the sum of the absolute values of its terms converges to a finite value. This means that no matter how the terms are rearranged, the series will still converge to the same value.

## 2. How is the absolute convergence of a series determined?

The absolute convergence of a series can be determined by taking the absolute value of each term and then applying any convergence tests, such as the ratio test or the root test. If the resulting series converges, then the original series is absolutely convergent.

## 3. What is the significance of absolute convergence in series analysis?

Absolute convergence is important because it guarantees that the series will converge to a unique value, regardless of the order in which the terms are added. This allows for easier manipulation and calculation of the series.

## 4. Can a series be absolutely convergent but not convergent?

No, if a series is absolutely convergent, it must also be convergent. This is because the absolute value of a term can only make the series converge faster or stay the same, but it cannot make it diverge.

## 5. How does absolute convergence differ from conditional convergence?

Conditional convergence occurs when a series converges, but its absolute value series diverges. This means that the series only converges if the terms are added in a specific order. In contrast, absolute convergence guarantees convergence regardless of the order of the terms.

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