Product of two series diverges

In summary, by defining the sequences an=(1/n) if n is even and 1/n^2 if n is odd, we can satisfy the conditions an>0 and an→0, Ʃ bn is bounded, but Ʃanbn diverges. This is achieved by multiplying the bounded sequence (-1)^n by the convergent sequence (-1)^n/n, which is not monotone in order to fail Abel's Test and the Alternating Series Test. While this may be a new way of defining sequences, it effectively solves the problem at hand.
  • #1
the_student
6
0

Homework Statement


Find sequences an and bn such that: an>0 and an→0, Ʃ bn is bounded, but Ʃanbn diverges.


The Attempt at a Solution


The idea is that bn should be -1^n or -1^(n+1) and when multiplied by an the odd (larger) terms of the new sequence diverge and overpower the smaller terms. Every sequence an I have tried still ends up converging (for example 1/n → 0 and diverges but (-1^n)/n converges.
 
Physics news on Phys.org
  • #2
What exactly are you trying that doesn't work? 1/n does diverge. Doesn't that suggest something?
 
  • #3
Yes, 1/n does diverge, however when multiplied by the bounded sequence (-1^n) you get (-1^n)/n which converges. It seems like I am missing the point, but I have focused on 1/n for some time and can't come up with how I can get the product to behave as 1/n does.
 
  • #4
I also played with the series Ʃ(1/k^2 - 1/k) which diverges but couldn't figure out a way for it to be the product of required an and bn.
 
  • #5
the_student said:
I also played with the series Ʃ(1/k^2 - 1/k) which diverges but couldn't figure out a way for it to be the product of required an and bn.

What happens if you multiply the bounded sequence (-1)^n by the convergent sequence (-1)^n/n?
 
  • #6
That would work perfectly but the requirement an > 0 for all n isn't satisfied with that sequence.
 
  • #7
the_student said:
That would work perfectly but the requirement an > 0 for all n isn't satisfied with that sequence.

Good point. You are paying attention. Suppose you define an=(1/n) if n is even and 1/n^2 if n is odd? Pick an to be not monotone.
 
Last edited:
  • #8
Well, I do precisely need an to be not monotone so that it fails Abel's Test and the Alternating Series Test, but I am a little concerned about defining an the way you mentioned. It seems like it would work, but we have never defined any of our series that way even in the in class examples.
 
  • #9
the_student said:
Well, I do precisely need an to be not monotone so that it fails Abel's Test and the Alternating Series Test, but I am a little concerned about defining an the way you mentioned. It seems like it would work, but we have never defined any of our series that way even in the in class examples.

Time to start a new trend then. Can you show it works?
 
  • #10
Thanks for the help. I feel better about it now that I know I was mostly there except for basically how to write my thoughts down.
 

1. What does it mean for a product of two series to diverge?

When a product of two series diverges, it means that the product of the two series does not have a finite or well-defined value. In other words, the terms in the series grow without bound, resulting in an infinite product.

2. What are the conditions for the product of two series to diverge?

In order for the product of two series to diverge, at least one of the series must diverge and the other must not converge to zero. This can also occur if both series have terms that do not approach zero, or if one series has terms that alternate between positive and negative values.

3. How is the divergence of a product of two series different from the divergence of a sum of two series?

When a sum of two series diverges, it means that the terms in the series grow without bound. However, when a product of two series diverges, the terms in the series do not simply grow without bound, but rather they multiply together and result in an infinite product. This is a key difference between the two types of divergences.

4. Can a product of two convergent series also diverge?

No, a product of two convergent series cannot diverge. When both series converge, their product will also converge to a finite value. This can be seen through the use of the Cauchy product, which states that the product of two convergent series will also converge to a finite value.

5. How is the divergence of a product of two series related to other mathematical concepts such as limits and continuity?

The divergence of a product of two series is closely related to the concept of limits and continuity. In order for a product of two series to converge, the limits of both series must exist and be finite. Additionally, the product of two continuous functions will also be continuous, so if the two series are made up of continuous functions, then their product will also be continuous.

Similar threads

Doubt regarding the series ##\sum [\sqrt{n+1} - \sqrt{n}\,]##
    • Calculus and Beyond Homework Help
Replies
17
Views
1K
How to show that these sequences are summable?
    • Calculus and Beyond Homework Help
Replies
1
Views
255
Is Using the Comparison Test to Prove Divergence Valid for This Series?
    • Calculus and Beyond Homework Help
Replies
29
Views
1K
Convergence of oscillatory/geometric series
    • Calculus and Beyond Homework Help
Replies
1
Views
781
Why the series is divergent based on the Preliminary test
    • Calculus and Beyond Homework Help
Replies
2
Views
814
A problem with the convergence of a series
    • Calculus and Beyond Homework Help
Replies
5
Views
990
Show Picard iteration diverges
    • Calculus and Beyond Homework Help
Replies
2
Views
496
Accumulation points and divergence
    • Calculus and Beyond Homework Help
Replies
5
Views
1K
Is this series divergent or convergent?
    • Calculus and Beyond Homework Help
Replies
16
Views
1K
On pointwise convergence of Fourier series
    • Calculus and Beyond Homework Help
Replies
3
Views
413

Hot Threads

Recent Insights

Back
Top