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You should know that $\displaystyle e^z = \sum_{z = 0}^{\infty}\frac{z^n}{n!}$.I am trying to prove e^z converges on all C. Here is my attempt.
e^z=series(z^n/n!)
use the ratio test on the coefficents 1/n! gives lim(1/(n+1))=0, which from rudin means radius of convergence R=1/0 -> R=infinity.
The ratio test applies to the terms not the coefficients.I am trying to prove e^z converges on all C. Here is my attempt.
e^z=series(z^n/n!)
use the ratio test on the coefficents 1/n! gives lim(1/(n+1))=0, which from rudin means radius of convergence R=1/0 -> R=infinity.
Yes you are correct. The ratio test give the radius of convergence whenever the limit exists. (http://en.wikipedia.org/wiki/Power_series#Radius_of_convergence) The root test also give the radius of convergence. (http://en.wikipedia.org/wiki/Cauchy–Hadamard_theorem)In the context of radius of convergence R must be greater than or equal to 0. I'm pretty convinced that, for power series, whener the ratio/root test of co-efficents gives 0, then we have infinite radius of convergence. Anyone who cares to contradict that is free to give a counter example.