Analysis of an absolutely convergence of series

[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!

Homework Equations

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

 

[member 587159]

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]

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 $n > m \ge N$

 

[member 587159]

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?

 

[JD_PM]

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]

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)

 

[member 587159]

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.

 

[JD_PM]

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?

 

[JD_PM]

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 $2B$ ?