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Math Help - Radius of Convergence

  1. #1
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    Radius of Convergence

    Hello i need to determine the radius of convergense for the following series:

    \sum_{n=0}^{\infty}\frac{1}{\sqrt{n!}}x^n

    Im thinking the ratio test, but as im pretty uncertain about these types of problems, i'd like a second opinion, and a bit of help
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  2. #2
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    Yes, for any "radius of convergence" problem, the best way to go is either the ratio test or the root test (and most of the time the ratio test is much simpler than the root test).

    Here, a_n= \frac{1}{\sqrt{n!}}x^n so a_{(n+1)}= \frac{1}{\sqrt{n+1}}x^{n+1}.

    Then \left|\frac{a_{n}}{a_{n+1}}\right|= \frac{\sqrt{n!}}{\sqrt{(n+1)!}}|x| = \sqrt{\frac{n!}{(n+ 1)!}}|x|

    What is the limit of that as n goes to infinity?
    For what x is that limit less than 1?
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    The limit is 1, so x < 1 will make the limit less than 1 making the radius of convergence = 1 ?
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    Quote Originally Posted by Zaph View Post
    The limit is 1, so x < 1 will make the limit less than 1 making the radius of convergence = 1 ?
    note that \frac{n!}{(n+1)!} = \frac{1}{n+1} ... care to rethink your response?
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    Quote Originally Posted by skeeter View Post
    note that \frac{n!}{(n+1)!} = \frac{1}{n+1} ... care to rethink your response?
    hmm, \frac{1}{n+1} diverges right?.. Im lost
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    Quote Originally Posted by Zaph View Post
    hmm, \frac{1}{n+1} diverges right?.. Im lost
    \frac{1}{n + 1} \to \frac{1}{\infty} \to 0 as n \to \infty.


    So what do you think the radius of convergence has to be?
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    Aha! So the radius of convergence is infinity or R? since the series goes to 0.
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    Quote Originally Posted by Zaph View Post
    Aha! So the radius of convergence is infinity or R? since the series goes to 0.
    Yes, since this limit is 0, it is always <1.

    So the series converges for all x.
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  9. #9
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    Thanx a bunch everyone, im really struggling with these types of problems, but its getting better and better
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