Difficult(?) convergence problem

  • Thread starter Thread starter Atropos
  • Start date Start date
  • Tags Tags
    Convergence
Atropos
Messages
9
Reaction score
0

Homework Statement




Show that if \vartheta is any constant not equal to 0 or a multiple of 2\pi, and if u_{0}, u_{1}, u_{2} is a series that converges monotonically to 0, then the series \sum u_{n} cos(n\vartheta +a) is also convergent, where a is an arbitrary constant.



Homework Equations





The Attempt at a Solution



I have attempted to show convergence via Cauchy's root test, Dirichlet's test, and Abel's test. All 3 of these attempts were unsucessful as one or more conditions required for the tests was not met.
 
Physics news on Phys.org
I have attempted to show convergence via Cauchy's root test, Dirichlet's test, and Abel's test. All 3 of these attempts were unsucessful as one or more conditions required for the tests was not met.

What are the objections to the Dirichlet test?

I'd think that | \sum_{i=1}^n (cos(n\vartheta)| would be bounded since a run of positive terms is followed by a run of negative terms. Likewise for sin(n\vartheta).

What's the longest run of postive terms that can happen? For \vartheta > 0 there is some smallest m so M \vartheta > 2 \pi Intuitively, I'd think 2M would be plenty big to bound it.
 
Wow, I'm an idiot.

I was so hung up on the cosine part of the sum that I completely forgot about the monotonic series u_n. I was only paying attention to the fact that cosine was sinusoidal ad therefore f_n>f_n+1>0 couldn't apply.

...but it does apply to u_n.

Thank you for pointing that out. :biggrin:
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Replies
3
Views
1K
Replies
4
Views
1K
Replies
5
Views
2K
Replies
8
Views
2K
Replies
1
Views
1K
Replies
11
Views
3K
Replies
24
Views
3K
Back
Top