Find Radius of Convergence of this Complex Series

by Ad123q
Tags: complex, convergence, radius, series
Ad123q is offline
Apr24-10, 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.
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mathman is offline
Apr24-10, 03:40 PM
Sci Advisor
P: 5,935
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.
Ad123q is offline
Apr25-10, 06:58 AM
P: 19
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.

mathman is offline
Apr25-10, 04:26 PM
Sci Advisor
P: 5,935

Find Radius of Convergence of this Complex Series

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.

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