Convergence/Divergence of Tricky Series: Tips and Tricks for Ʃn=1∞ in/n

In summary: Thanks for checking!In summary, the author is trying to determine if the series Ʃn=1∞ in/n converges or diverges and is having difficulty doing so. The author has tried different tests and has not been able to determine convergence. If convergence did exist, it would only be on the boundary z=1.
  • #1
matt.qmar
7
0
Hey,

I am trying to determine the convergence/divergence of

Ʃn=1 in/n.

I have tried all the tests I could think of (Comparison, Ratio, Root, nth term) and cannot determine it's convergence.

If there was a formula, for say, the Mth partial sum SM then, if the limit as M → ∞ of SM is L, we have convergence to L but I can't seem to arrange for the thing to add up the first M terms.

Clearly, I think something can be done with the fact that in = {i, -1, -i, 1} repeatably with periodicity 4. I'm not sure how this can exactly be of help though!

Any help appreciated, thanks.
 
Physics news on Phys.org
  • #2
Do you know the power series for [itex]\log(1-z)[/itex]? It has a raidus of convergence equal to one. However convergence on the boundary [itex]|z|=1[/itex] is well, what? Take a look at Wikipedia for "Radius of Convergence".
 
  • #3
Ah! I see, and we have |i| = 1 but i ≠ 1 so there we go!

Thanks a bunch. Factoring out the -1 seems to make it all go quite nicely.
 
  • #4
Ʃn=1 in/n = -log(1-i), woo hoo!
 
  • #5
You know, that's really not good enough. Why does it converge on the boundary except for z=1? Or rather prove that it does. If you wish anyway.
 
  • #6
Ah, Thanks for the heads up. I suppose I sort of took that for granted. Would http://en.wikipedia.org/wiki/Abel's_test#Abel.27s_test_in_complex_analysis works to show convergence for all z ≠ 1 on the boundary (in particular, at our point i ≠ 1), as the an's are monotonically decreasing? Some the an, however, are negative...

Trying to take some zo ≠ 1 with |zo| = 1 and using some other test (root, comparison, etc.) doesn't seem to work either, though...
 
  • #7
Yeah, I don't know off-hand how to show that. There are tests to check for convergence on the boundary. Would need to look into thoses. Don't thinik Abel test would apply though.
 

1. What is a "Tricky Complex Series"?

A tricky complex series is a mathematical concept that involves a sequence of complex numbers, where each term is dependent on the previous term. These series can be difficult to analyze and often require advanced mathematical techniques to determine their convergence or divergence.

2. How do you determine the convergence of a "Tricky Complex Series"?

The convergence of a tricky complex series is typically determined by using a combination of techniques such as the ratio test, root test, or comparison test. These tests involve evaluating the limit of the series as the number of terms approaches infinity, and can be used to determine if the series converges or diverges.

3. What makes a "Tricky Complex Series" different from other series?

A tricky complex series differs from other series in that it involves complex numbers, which have both a real and imaginary component. This adds an extra level of complexity to the analysis of the series, as the convergence of the series can be affected by the values of both the real and imaginary parts of the complex numbers.

4. Are there any real-world applications of "Tricky Complex Series"?

Yes, tricky complex series have many real-world applications in fields such as physics, engineering, and economics. For example, they can be used to model the behavior of electrical circuits, fluid flow, or financial markets.

5. What are some strategies for solving a "Tricky Complex Series"?

Some strategies for solving a tricky complex series include using techniques such as partial fraction decomposition, algebraic manipulation, and geometric series. It is also helpful to have a strong understanding of complex numbers and their properties, as well as a working knowledge of convergence tests.

Similar threads

Replies
15
Views
2K
Replies
11
Views
2K
Replies
3
Views
898
Replies
6
Views
525
Replies
14
Views
2K
Replies
3
Views
886
  • Calculus
Replies
3
Views
869
  • Calculus and Beyond Homework Help
Replies
2
Views
87
Replies
9
Views
2K
  • Calculus
Replies
1
Views
1K
Back
Top