What is the radius of convergence of

  • #1
1,661
2

Homework Statement



z ∈ ℂ

What is the radius of convergence of (n=0 to ∞) Σ anzn?

Homework Equations



I used the Cauchy-Hardamard Theorem and found the lim sup of the convergent subsequences.

[tex]a_n = \frac{n+(-1)^n}{n^2}[/tex]

limn→∞ |an|1/n

The Attempt at a Solution



I think that the radius of convergence is one, i.e. |z| < 1. I figured that the numerator would tend to n with the oscillating 1 and so you'd get [tex]\frac{n^{1/n}}{n^{2/n}} = 1[/tex]

R = 1/limn→∞ |an|1/n
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,260
619

Homework Statement



z ∈ ℂ

What is the radius of convergence of (n=0 to ∞) Σ anzn?

Homework Equations



I used the Cauchy-Hardamard Theorem and found the lim sup of the convergent subsequences.

[tex]a_n = \frac{n+(-1)^n}{n^2}[/tex]

limn→∞ |an|1/n

The Attempt at a Solution



I think that the radius of convergence is one, i.e. |z| < 1. I figured that the numerator would tend to n with the oscillating 1 and so you'd get [tex]\frac{n^{1/n}}{n^{2/n}} = 1[/tex]

R = 1/limn→∞ |an|1/n
That's a little informal but it looks fine. You might want worry about what happens at z=1 and z=(-1) if you are concerned about the boundary cases.
 
  • #3
1,661
2
That's a little informal but it looks fine. You might want worry about what happens at z=1 and z=(-1) if you are concerned about the boundary cases.
I assume a singularity is there and outside of the disk the series is divergent. Well, I'm leaving out a few bits of information in my "proof" here. I'll state a bit more of the background. What would a more formal proof look like?
 
  • #4
Dick
Science Advisor
Homework Helper
26,260
619
I assume a singularity is there and outside of the disk the series is divergent. Well, I'm leaving out a few bits of information in my "proof" here. I'll state a bit more of the background. What would a more formal proof look like?
Rather than just ignoring the (-1)^n handle it with a squeeze type thing. E.g. n/2<=n+(-1)^n<=2n. You can easily find the outer limits. Discuss what happens when z=1 or z=(-1).
 

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