# Complex Analysis Radius of Convergence.

• I
Hello, I have two questions regarding the Radius of convergence.

1. What should we do at the interval (R-eps, R)
2. It used definition to prove radius of convergence, but I am not sure if it is necessary-sufficient condition of convergence. I get that this can be a sufficient condition but not sure of the necessity

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• 스크린샷 2016-06-21 오후 4.01.05.png
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If I understand the question, this may help -- The ε was picked arbitrarily small so the proof shows convergence all the way to ε = 0. So there really is no specific interval (R-ε, R) to worry about.

what I am concerend about is,

hm the author used (L+eps)|Z|<1 to argue that L^n*|Z|^n < (L+eps)^n|Z|^n < r^n whereas r<1

but the thing is as I showed at the uploaded png file (if the derivation is correct), it is clear that |Z|= R-eps satisfies the condition |Z| < R and if |Z|= R-eps, r = 1.

which implies (L+eps)|Z|<1 cannot be true. for all |Z|<R
Ah, also I made a mistake. It should be R+R^2*eps-eps-R*eps^2

hmm do you agree?

Last edited:
If I understand the question, this may help -- The ε was picked arbitrarily small so the proof shows convergence all the way to ε = 0. So there really is no specific interval (R-ε, R) to worry about.

maybe I screwed up at somewhere.

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maybe I screwed up at somewhere.
Maybe not. But the point is that ε >0 is arbitrarily small (or r is arbitrarily near 1) so the the interval you are worried about shrinks down to nothing.

Maybe not. But the point is that ε >0 is arbitrarily small (or r is arbitrarily near 1) so the the interval you are worried about shrinks down to nothing.

Although we ignore the interval, the argument that (L+eps)|Z|<1 does not hold true if we take |Z| = R - eps. (I guess) It just bugs me and I am stuck at that page ;(

and my last conclusion R<1 doesn't seem right at all too.

Stephen Tashi