Is There a Counterexample for Convergence of bn Series?

[TheForumLord]

Homework Statement

Let an be a series of positive elements such as lim_n->infinity_an=0 and the series Sigma_n=1-infinity_an diverge.
Let bn=(-1)^n*an
Prove or find counterexamples for the next propositions:
1. 1. Sigma_bn diverge.
2. Sigma_bn doesn't absolutely converge.
3. Sigma_bn converge.

Homework Equations

The Attempt at a Solution

I need a counterexample for 3...I have a counterexample for 1 and I think I can prove 2...
3 is all I need
TNX everyone!

 

[Dick]

Pick your favorite series a_n satisfying the premises. Can you think of a way to create a new series where all of the original a_n terms occupy positions in the new series with even indices? For example.

 

[TheForumLord]

Hmmm... I thought about an=1/n... I can't think of any possible way to create from this series a new series that will satisfy the results of part 3...
Will you help me? :)

TNX

 

[Dick]

Ok, take 1/n. Create a new series by putting all of the entries of 1/n into the even slots of the new series. Put something harmless into the odd slots. You want (-1)^n*an to diverge, right?

 

[TheForumLord]

Wow you're right... We can build a series of this form:
an=[1/n if n is even] & [0 if n is odd] ... Then obviously the series Sigma_an diverge& an->0 as needed...But the series bn is excatly the series an so it's also diverge!

AM I right?
TNX a lot!

 

[Dick]

Sure, {0,1/2,0,1/4,0,1/6,...} works, which seems to be what you are describing. I was thinking of {0,1,0,1/2,0,1/3,...}. Take your pick.

 

[TheForumLord]

TNX a lot!
The series I meant was {0,1,0,1/2,0,1/3,...} but my wording wasn't exact...

 

[vela]

That series doesn't strictly meet the requirements of a_n since the original problem says that a_n is a positive sequence, not a non-negative one.

 

[Dick]

That series doesn't strictly meet the requirements of a_n since the original problem says that a_n is a positive sequence, not a non-negative one.

Very observant!

 

[TheForumLord]

Wow...That's right...So, What can we do about it?

TNX

 

[vela]

The problem probably has to do with the convergence theorem for alternating series. If you want one that doesn't converge, look for series that doesn't meet the conditions required of the theorem.

 

[TheForumLord]

I need to find a series that doesn't descend. But I can't think of any non-descending series with positive elements such as lim_an_n->infiniy=0 ...

CAn you help me please?
TNX

 

[vela]

The convergence test requires a monotonically decreasing sequence. That means you need a_k > a_{k+1} for all k>N for some N. If you have a sequence that bounces up and down, it won't meet the condition even if the limit is going to zero. For example, in your earlier solution, you filled the odd-numbered spots with 1/n. Can you think of another sequence whose limit is zero but where the terms are bigger than 1/n? Use that sequence to take the place of the zeros. Of course, if you accomplish this, you'll only know that the series doesn't converge; you won't know if it diverges.

Perhaps a better way of looking at it would be as a sum of differences:

S = (a_0-a_1) + (a_2-a_3) + (a_4-a_5) + ...

If you can figure out how to make the differences into, say, the harmonic series, you'd have a series that you know diverges.

 

[Dick]

Wow...That's right...So, What can we do about it?

TNX

You really can't think of anything to do??! I just suggested you put something 'harmless' in the odd locations. The 'harmless' thing doesn't have to zeros. It could be positive stuff. This whole issue is just a technicality.

 

[TheForumLord]

Very nice guide! Tnx a lot...I think I found a counterexample:
Let's define the series an :
an=1/0.5n if n is even and an=1/n^2 if n is odd...
We'll get the series {1,1,1/9,1/2,1/25,1/3,1/49,1/4,...Etc...} It's pretty obvious that
an->0 as n->infinity but the series Sigma_an diverges as needed... We can also see that the series Sigma_(-1)^n*an diverges because:
Sigma_bn = -1+1-1/9 + 1/2 -1/25 + 1/3 - 1/49 + 1/4 +...

Am I wrong or something?

Another possible counterexample can be (I figured it out after I've read your guide) :
an = 1/n when n is odd and an = 2/n when n is even ... We'll get a(n+1)-an = 1/n and it diverges Hence bn diverges...

Verification is needed!

TNX a lot!

 

[Dick]

Very nice guide! Tnx a lot...I think I found a counterexample:
Let's define the series an :
an=1/0.5n if n is even and an=1/n^2 if n is odd...
We'll get the series {1,1,1/9,1/2,1/25,1/3,1/49,1/4,...Etc...} It's pretty obvious that
an->0 as n->infinity but the series Sigma_an diverges as needed... We can also see that the series Sigma_(-1)^n*an diverges because:
Sigma_bn = -1+1-1/9 + 1/2 -1/25 + 1/3 - 1/49 + 1/4 +...

Am I wrong or something?

Another possible counterexample can be (I figured it out after I've read your guide) :
an = 1/n when n is odd and an = 2/n when n is even ... We'll get a(n+1)-an = 1/n and it diverges Hence bn diverges...

Verification is needed!

TNX a lot!

Sure. The first one is fine. As for the last one, you'd better write out a few terms. Your description isn't very clear.

 

[TheForumLord]

Sure...My description was very vague indeed.
I meant this series:
{1/2 , 1 , 1/3, 2/3, 1/4, 1/2 , 1/5, 2/5 ,...}
Hence:
In the odd entries we put the series 1/n from n=2 to infinity in the same order as the series itself and not according to the n's of the entries...
In the even entries we put the series 2/n from n=1 to infinity...
It will eventually diverge ofcourse...

TNX