Radius and interval of convergence for

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Homework Statement



Hey all. I'm being asked to find the radius and interval of convergence for the series from 1 to infinity:

x^n/(1*3*5...(2n-1))

I have a feeling this is pretty straight forward (just apply the ratio test etc...), but my trouble lies in defining the denominator. Something to do with factorials? How do I get something to represent 1*3*5 all the way to *(2n-1)?
 

Answers and Replies

  • #2
Dick
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Ok, so a_n=1/(1*3*5...(2n-1)). I think that represents it pretty well. What's a_(n+1)/a_n?
 
  • #3
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a_(n+1)/a_n would be =

1*3*5...(2n-1) / 1*3*5...(2n-1)(2n+1)

correct? In this case it simplifies to 1/(2n+1)?
 
  • #4
Dick
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a_(n+1)/a_n would be =

1*3*5...(2n-1) / 1*3*5...(2n-1)(2n+1)

correct? In this case it simplifies to 1/(2n+1)?
Absolutely correct. So what does the ratio test tell you?
 
  • #5
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Convergent for all x^n I believe, so radius of convergence = inf?
 
  • #6
Dick
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Convergent for all x^n I believe, so radius of convergence = inf?
Sure. lim a_(n+1)/a_n=0. So infinite radius of convergence. Converges for all x.
 
  • #7
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Thanks very much!
 
  • #8
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Just to nitpick on the absolutely perfect argument above --- it's technically an implication of the ratio test and not because of the ratio test. If the power series is of the form [tex]\sum a_n x^n[/tex], then we define [tex]\beta = \limsup |a_n|^{1/n}[/tex] and the radius of convergence as [tex]R = 1 / \beta[/tex].

So, technically, one is supposed to compute [tex]\beta[/tex], but often it is hard to compute the n-th root of things. But the trick is that we see this inequality relationship
[tex]\liminf |a_{n+1} / a_n| \leq \liminf |a_n|^{1/n} \leq \limsup |a_n|^{1/n} \leq \limsup|a_{n+1} / a_n|[/tex]. And since if [tex]\lim |a_{n+1} / a_n|[/tex] exists, then [tex]\lim |a_{n+1} / a_n| = \liminf |a_{n+1} / a_n| = \limsup|a_{n+1} / a_n|[/tex], implying immediately that [tex]\limsup |a_n|^{1/n} = \lim |a_{n+1} / a_n| [/tex].

I'm just nitpicking :P
 

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