Absolute convergence, boundedness, and multiplication

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Homework Help Overview

The discussion revolves around the convergence of series, specifically examining the conditions under which the series \(\sum_{n=1}^{\infty}x_ny_n\) converges given that \(\sum_{n=1}^{\infty}x_n\) converges absolutely and the sequence \(y_n\) is bounded.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the implications of absolute convergence and bounded sequences, questioning the validity of certain inequalities and the necessity of considering the signs of the terms involved. Some suggest integrating absolute values into the reasoning to clarify convergence.

Discussion Status

The discussion is active, with participants raising concerns about the initial approach and suggesting the need to incorporate absolute values. There is recognition of the importance of absolute convergence in specific scenarios, and some participants are attempting to reformulate their reasoning based on these insights.

Contextual Notes

There is an ongoing debate about the nature of the sequences involved, particularly regarding the alternating harmonic series and its properties. Participants are also considering the implications of boundedness in relation to convergence.

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Homework Statement


If the series [itex]\sum_{n=1}^{\infty}x_n[/itex] converges absolutely, and the sequence [itex](y_n)_n[/itex] is bounded, then the series [itex]\sum_{n=1}^{\infty}x_ny_n[/itex] converges.

Homework Equations


Definitions and theorems relating to series and convergence.

The Attempt at a Solution


If the sequence [itex]y_n[/itex] is bounded, then [itex]|y_n|\leq M[/itex] for all n. Then [itex]\sum_{k=1}^{n}x_ky_k\leq\sum_{k=1}^{n}Mx_k[/itex]. But [itex]M\sum_{n=1}^{\infty}x_n[/itex] is convergent which implies that [itex]\sum_{n=1}^{\infty}x_ny_n[/itex] is also by comparison. [itex]\blacksquare[/itex]

Now I believe this is the right path, however, I did not use the absolute convergence in this proof, and I didn't really consider if the terms in [itex]y_n[/itex] were negative... I need some help to know how to include these aspects.
 
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[itex]\sum_{k=1}^{n}x_ky_k\leq\sum_{k=1}^{n}Mx_k[/itex]

This isn't necessarily true. If both x and y alternate, say yn = -1n and xn = -1n / n, then it doesn't converge, even though y is bounded and the x-series converges. That is the step in which you need to integrate absolute convergence.
 
Sethric said:
[itex]\sum_{k=1}^{n}x_ky_k\leq\sum_{k=1}^{n}Mx_k[/itex]

This isn't necessarily true. If both x and y alternate, say yn = -1n and xn = -1n / n, then it doesn't converge, even though y is bounded and the x-series converges. That is the step in which you need to integrate absolute convergence.

I agree I need to integrate in absolute convergence, however [itex]x_n=(-1)^n/n[/itex] could not be since it is absolutely convergent, and the alternating harmonic series is not. But I get what you are getting at in that I must find a way to include the absolute convergence for particularly sticky situations.
 
[tex]\sum_{n=1}^\infty \frac{(-1)^n}{n}[/tex] is not absolutely convergent.

and the alternating harmonic series is not
That technically is an alternating harmonic series.

What I am driving at is using absolute values. Can you think of a way to do that step using absolute values that takes advantage of the absolute convergence of xn?
 
Sethric said:
[tex]\sum_{n=1}^\infty \frac{(-1)^n}{n}[/tex] is not absolutely convergent.


That technically is an alternating harmonic series.

What I am driving at is using absolute values. Can you think of a way to do that step using absolute values that takes advantage of the absolute convergence of xn?

Ah! I was trying to say that it is not absolutely convergent, so [itex]x_n[/itex] could not be what you had in your example. But you posted before I could edit it

I'll try and think a bit more about it but that has been what I have been trying to think of...
 
Let me try this:

You know the sequence yn is bounded. What do you know about:

[tex]\sum_{k=1}^\infty \left| x_k y_k \right|[/tex]

?
 
Sethric said:
Let me try this:

You know the sequence yn is bounded. What do you know about:

[tex]\sum_{k=1}^\infty \left| x_k y_k \right|[/tex]

?

Okay, I think I got it:

Given [itex]\sum_{n=1}^{\infty}x_n[/itex] converges absolutely, and [itex]y_n[/itex] is bounded, then [itex]|y_n|\leq M[/itex] for all [itex]n[/itex] for some [itex]M>0[/itex]. Consider

[tex]M\sum_{k=1}^{n}|x_k|\geq\sum_{k=1}^{n}|x_k||y_k|=\sum_{k=1}^{n}|x_ky_k|.[/tex]

But [itex]M\sum_{k=1}^{n}|x_k|[/itex] is convergent which implies that [itex]\sum_{k=1}^{n}|x_ky_k|[/itex] is also convergent. But if the sequence of partial sums converge, so does the series. Then since the series is absolutely convergent, it is convergent. [itex]\blacksquare[/itex]
 

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