Find Radius of Convergence of this Complex Seriesby Ad123q Tags: complex, convergence, radius, series 

#1
Apr2410, 03:06 PM

P: 19

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. 



#2
Apr2410, 03:40 PM

Sci Advisor
P: 5,935

The ratio test should work. The ratio of the n term to the (n1) term is (ze)^{n}.
If ze < 1, the ratio > 0, while if ze > 1, tie ratio becomes infinite. 



#3
Apr2510, 06:58 AM

P: 19

Thanks, though I'm still not 100% sure about how to get this.
Using ratio test I obtain mod[ (ze)^(n+1)! / (ze)^n! ] = mod[ (ze)^n!(n+11) ] = mod[ (ze)^n.n! ] Then I'm not sure where to go from here. 



#4
Apr2510, 04:26 PM

Sci Advisor
P: 5,935

Find Radius of Convergence of this Complex Series
mod[ (ze)^n.n! ] = [mod(ze)]^n.n!
When mod(ze) < 1, the large exponent makes the expression very small, while if mod(xe) > 1, it makes the expression very large. Therefore the radius of convergence = 1. 


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