Proving Convergence of e^z Series Using Ratio Test on Coefficients

[Poirot1]

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.

 

[Prove It]

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.

You should know that $\displaystyle e^z = \sum_{z = 0}^{\infty}\frac{z^n}{n!}$.

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.

 

[CaptainBlack]

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.

CB

 

[Poirot1]

Yes you can do it that way but you can also do the test on the coefficents then let R be the reciprocal. I have done this before so I know it works, I was just wondering when you get 0 can you just say 1/0 = infinity?

 

[HallsofIvy]

No, you can't "just say" that. However, it is fairly easy to prove that if $a_n$ converges to 0 then $\frac{1}{a_n}$ does not converge. If you add that $a_n> 0$ for all n, then if diverges to $+\infty$.

 

[Poirot1]

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.

 

[Sudharaka]

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.

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)