# 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

## Answers and Replies

FactChecker
Science Advisor
Gold Member
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:
FactChecker
Science Advisor
Gold Member
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
Science Advisor
It isn't clear to me what is to be proven. We've only been shown the last part of the text that states "the theorem".