• Support PF! Buy your school textbooks, materials and every day products Here!

Complex series

  • Thread starter ehrenfest
  • Start date
1,996
1
1. Homework Statement
Investigate the behavior (convergence or divergence) of [itex]\sum_n 1/(1+z^n)[/itex] where z is complex.



2. Homework Equations



3. The Attempt at a Solution
If the modulus of z is less than 1, it is not hard to show that the limit of the sequence is not 0 (it is actually not finite) and thus the series cannot converge. But if the modulus of z is greater than or equal to 1, I don't what to apply. The root test? The ratio test? The comparison test?
 

Answers and Replies

1,356
0
Would you care to show how the series diverges if |z| < 1?

I determined that if |z| > 1, then series converges. I did this by comparing |zn| and |1 + zn| and using the comparison test.
 
1,996
1
Would you care to show how the series diverges if |z| < 1?

I determined that if |z| > 1, then series converges. I did this by comparing |zn| and |1 + zn| and using the comparison test.
I highly doubt that would work because I used that comparison to show it diverges when |z| < 1. Please explain exactly how you compared the two series.
 
229
0
I highly doubt that would work because I used that comparison to show it diverges when |z| < 1. Please explain exactly how you compared the two series.
why don't you show us what you did? it converges for |z| > 1
 
1,356
0
I highly doubt that would work because I used that comparison to show it diverges when |z| < 1. Please explain exactly how you compared the two series.
I will, once I see exactly how you showed that it diverges for |z| < 1.
 
1,996
1
[tex] |\frac{1}{1+z^n}| \geq \frac{1}{1+|z|^n} \geq 1/|z|^n [/tex]

So the sequence of terms doesn't even go to 0 when |z| < 1
 
Last edited:
229
0
[tex] |\frac{1}{1+z^n}| \geq \frac{1}{1+|z|^n} \geq 1/|z|^n [/tex]

So the sequence of terms doesn't even go to 0 when |z| < 1
this is wrong 1/(1 + |z|^n) >= 1/|z|^n
 
229
0
for |z| > 1, |1/(1 + z^n)| <= |1/z^n| = |1/z|^n and |1/z| < 1, so we have convergence
 
229
0
for |z| <= 1, suppose we had convergence, then lim n-> inf |1/(1 + z^n)| = 0, so lim 1/lim |1 + z^n| = 0, so lim |1 + z^n| = inf, but |1 + z^n| <= 1 + 1 = 2, a contradiction, so it cannot converge for |z| <= 1
 
1,356
0
I'm confused now. If Re(zn) < 0, then |zn| > |1 + zn| right? If Re(zn) >= 0, then |zn| < |1 + zn|. I think we have to consider both cases.
 
229
0
I'm confused now. If Re(zn) < 0, then |zn| > |1 + zn| right? If Re(zn) >= 0, then |zn| < |1 + zn|. I think we have to consider both cases.

edit: you're right in that I made a mistake. Take n = 1, z = -1 + i, then |z| = sqrt(2) and |1 + z| = 1, so |z| > |1 + z|, i forgot these are complex numbers here
 
Last edited:
1,356
0
for |z| <= 1, suppose we had convergence, then lim n-> inf |1/(1 + z^n)| = 0, so lim 1/lim |1 + z^n| = 0, so lim |1 + z^n| = inf, but |1 + z^n| <= 1 + 1 = 2, a contradiction, so it cannot converge for |z| <= 1
By the triangle inequality, |1 + zn| <= 1 + |z|n so

[tex]\frac{1}{1 + |z|^n} \le \frac{1}{|1 + z^n|} = \left|\frac{1}{1 + z^n}\right|[/tex]

and furthermore

[tex]\sum_{n=0}^\infty \frac{1}{1 + |z|^n} \le \sum_{n=0}^\infty \left|\frac{1}{1 + z^n}\right|[/tex]

So by the comparison test, if the LHS diverges, then so does the RHS. Now if |z| <= 1, then 1/2 <= 1/(1 + |z|n) <= 1 for all n and so the limit lies in [1/2, 1] and by the zero-test, the LHS diverges. Thus, so does the RHS.

The problem is this doesn't tell us if the original series of 1/(1 + zn) diverges.
 

Related Threads for: Complex series

  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
3
Views
1K
Replies
15
Views
2K
  • Last Post
Replies
0
Views
7K
  • Last Post
Replies
2
Views
602
  • Last Post
Replies
1
Views
1K
Top