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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 states that when you evaluate $\displaystyle \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|$, if this limit is less than 1, the series is convergent, if this limit is greater than 1, the series is divergent, and if the limit is 1, the test is inconclusive. Since you are trying to find the values for which this series is convergent, you need to set $\displaystyle \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| < 1$, simplify, and see what values of z will satisfy that inequality.

- Jan 26, 2012

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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.

CB

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- Jan 29, 2012

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- Feb 5, 2012

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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)