What is the radius of convergence of

Shackleford
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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.

a_n = \frac{n+(-1)^n}{n^2}

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 \frac{n^{1/n}}{n^{2/n}} = 1

R = 1/limn→∞ |an|1/n
 
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Shackleford said:

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.

a_n = \frac{n+(-1)^n}{n^2}

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 \frac{n^{1/n}}{n^{2/n}} = 1

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.
 
Dick said:
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?
 
Shackleford said:
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).
 
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...

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