November 15th, 2016, 06:04 PM  #1 
Member Joined: Aug 2016 From: illinois Posts: 46 Thanks: 0  Interval of Convergence
Im trying to find the interval of convergence of a series. I used the Ratio test for absolute convergence, but that limit ended up being zero. For the other problems i've done, it usually comes out to 1 and then i multiply that by X and find the I.O.C. Im not sure what that means when the limit is zero though.

November 15th, 2016, 06:38 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 3,101 Thanks: 1677 
maybe this is an example ... $\displaystyle \sum_{n=0}^\infty \dfrac{x^n}{n!}$ $\displaystyle \lim_{n \to \infty} \bigg\dfrac{x^{n+1}}{(n+1)!} \cdot \dfrac{n!}{x^n} \bigg$ $\displaystyle x \cdot \lim_{n \to \infty} \dfrac{1}{n+1} = 0$ Interval of convergence is $\infty <x< \infty$ 
November 15th, 2016, 07:09 PM  #3 
Member Joined: Aug 2016 From: illinois Posts: 46 Thanks: 0 
Ok thanks, but why is it infinity to infinity, i dont really get it.

November 15th, 2016, 07:35 PM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 3,101 Thanks: 1677 
$x \cdot 0 < 1$ ... for what values of x is that inequality true?
Last edited by skeeter; November 15th, 2016 at 07:38 PM. 
November 15th, 2016, 07:37 PM  #5 
Member Joined: Aug 2016 From: illinois Posts: 46 Thanks: 0 
Okay that makes sense now, thanks a lot.


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convergence, interval 
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