Investigating Complex Series: Convergence or Divergence?

In summary, the behavior of the series \sum_n 1/(1+z^n) where z is a complex number depends on the modulus of z. If the modulus of z is less than 1, the series diverges. If the modulus of z is greater than or equal to 1, the series converges. This can be shown by comparing the series to other known series and using the comparison test.
  • #1
ehrenfest
2,020
1

Homework Statement


Investigate the behavior (convergence or divergence) of [itex]\sum_n 1/(1+z^n)[/itex] where z is complex.

Homework Equations


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?
 
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  • #2
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.
 
  • #3
e(ho0n3 said:
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.
 
  • #4
ehrenfest said:
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
 
  • #5
ehrenfest said:
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.
 
  • #6
[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
 
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  • #7
ehrenfest said:
[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
 
  • #8
for |z| > 1, |1/(1 + z^n)| <= |1/z^n| = |1/z|^n and |1/z| < 1, so we have convergence
 
  • #9
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
 
  • #10
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.
 
  • #11
e(ho0n3 said:
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
 
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  • #12
ircdan said:
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.
 

1. What does it mean for a series to converge or diverge?

Convergence and divergence refer to the behavior of a series as its terms are added together. A convergent series is one in which the sum of its terms approaches a finite number as more terms are added. A divergent series is one in which the sum of its terms either grows infinitely large or oscillates between different values as more terms are added.

2. How do you determine if a series is convergent or divergent?

There are several methods for determining the convergence or divergence of a series, including the comparison test, the ratio test, and the integral test. These methods involve comparing the given series to a known series with known convergence or divergence properties. Additionally, the series may also have a known formula for its sum, which can be used to determine its convergence or divergence.

3. What is the importance of investigating the convergence or divergence of a series?

Understanding the convergence or divergence of a series is important in many areas of mathematics, physics, and engineering. In particular, it allows us to determine the behavior of infinite sums and to make predictions about the behavior of systems that involve such sums. This knowledge is also crucial in numerical analysis and computer science, where series are often used in algorithms and simulations.

4. Can a series be both convergent and divergent?

No, a series cannot be both convergent and divergent. A series must exhibit either convergence or divergence behavior, as these are mutually exclusive properties. However, it is possible for a series to be conditionally convergent, meaning that it is convergent but not absolutely convergent. In this case, the series still has a finite sum, but rearranging the terms of the series can lead to a different sum.

5. What are some real-world applications of investigating the convergence or divergence of a series?

The concept of series convergence and divergence has numerous applications in fields such as finance, physics, and engineering. For example, in finance, series are used to model the growth of investments and the payment of annuities. In physics, series are used to approximate solutions to differential equations and model physical systems. In engineering, series are used in signal processing and control systems. Additionally, the study of convergence and divergence has led to the development of important mathematical tools and techniques that are used in many areas of science and engineering.

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