May 17th, 2017, 03:21 AM  #1 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 211 Thanks: 2  Series Convergence
The set of all values of $a$ for which the series $\displaystyle \sum_{n=1}^{\infty} \frac {a^n }{n! }$ converges is A) $\displaystyle (0,\infty) $ B) $\displaystyle (\infty,0] $ C) $\displaystyle (\infty,\infty) $ D) $\displaystyle (1,1) $ I have two opinions for this question. Option D  As it is convergent at $0$ and also around $0$ for the given $n$ values. Option B As I know for $a= 1$ it is convergent but for the other values of $a$ like $2,3...$ does it converge ? Please clarify and also rectify if I'm wrong. Thank you 
May 17th, 2017, 04:04 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,692 Thanks: 1351 
using the ratio test for convergence ... $\displaystyle \lim_{n \to \infty} \left \dfrac{a^{n+1}}{(n+1)!} \cdot \dfrac{n!}{a^n} \right < 1$ $\displaystyle a \lim_{n \to \infty} \dfrac{1}{n+1} <1$ $a \cdot 0 < 1$ for all values of $a \implies$ interval of convergence is $(\infty,\infty)$ 
May 17th, 2017, 04:35 AM  #3  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 211 Thanks: 2  Quote:
I'm confused. Can you please explain where the series is convergent in which intervel ? and at which values of $a$ ? Thanks  
May 17th, 2017, 05:01 AM  #4  
Math Team Joined: Jul 2011 From: Texas Posts: 2,692 Thanks: 1351  Quote:
Watch the linked video to see another example of using the ratio test to determine an interval of convergence.  
May 17th, 2017, 06:10 AM  #5 
Math Team Joined: Jul 2011 From: Texas Posts: 2,692 Thanks: 1351 
I linked the wrong video ... meant to post this one. 
May 17th, 2017, 08:12 PM  #6 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 211 Thanks: 2  

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convergence, series 
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