Absolute convergence, boundedness, and multiplication

[jfy4]

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.

 

[Sethric]

[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 y_n = -1ⁿ and x_n = -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.

 

[jfy4]

[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 y_n = -1ⁿ and x_n = -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.

 

[Sethric]

[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 x_n?

 

[jfy4]

[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 x_n?

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...

 

[Sethric]

Let me try this:

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

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

?

 

[jfy4]

Let me try this:

You know the sequence y_n 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]