Find Radius of Convergence of this Complex Series

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Discussion Overview

The discussion revolves around finding the radius of convergence for the complex series defined as the sum from n=1 to infinity of (z - e)^n!. Participants explore the application of the Ratio Test and express uncertainty regarding the factorial in the exponent.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses uncertainty about using the Ratio Test for the series due to the factorial in the exponent.
  • Another participant suggests that the Ratio Test should work and provides a ratio of terms as (z-e)^n.
  • A third participant attempts to apply the Ratio Test but becomes confused about the manipulation of terms involving factorials.
  • A later reply discusses the behavior of the expression based on the modulus of (z-e) and concludes that the radius of convergence is 1 based on this reasoning.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the application of the Ratio Test, and there is ongoing uncertainty regarding the manipulation of terms involving factorials.

Contextual Notes

Some participants' reasoning depends on the interpretation of the factorial in the context of the Ratio Test, which remains unresolved. There are also missing steps in the mathematical manipulations presented.

Ad123q
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Hi, am a bit stuck with this.

Find the radius of convergence of the complex series

(Sigma n=1 to infinity) (z - e)^n!

I know that the answer is R=1 but I'm not sure how to get there.
It's the factorial as a power which I'm not sure about, have seen this in some other problems too.
I have tried using the Ratio Test to determine radius of convergence but doesn't seem to be working.

Any ideas anyone?

Thanks in advance, and apologies for the lack of Latex.
 
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The ratio test should work. The ratio of the n term to the (n-1) term is (z-e)n.
If |z-e| < 1, the ratio -> 0, while if |z-e| > 1, tie ratio becomes infinite.
 
Thanks, though I'm still not 100% sure about how to get this.

Using ratio test I obtain

mod[ (z-e)^(n+1)! / (z-e)^n! ]
= mod[ (z-e)^n!(n+1-1) ]
= mod[ (z-e)^n.n! ]

Then I'm not sure where to go from here.
 
mod[ (z-e)^n.n! ] = [mod(z-e)]^n.n!

When mod(z-e) < 1, the large exponent makes the expression very small, while if mod(x-e) > 1, it makes the expression very large. Therefore the radius of convergence = 1.
 
Last edited:

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