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Math Help - Finding values for which a sequence converges..

  1. #1
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    Finding values for which a sequence converges..

    For which values of x does \sum_{n=0}^\infty \frac{x^{2n}}{\sqrt{n+1}} converge? Justify your answer.

    Now, I reckon the sequence converges for -1>x<1, because anything else looks like it will just diverge... but how do I go about "justifying" this?
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  2. #2
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    Quote Originally Posted by Unenlightened View Post
    For which values of x does \sum_{n=0}^\infty \frac{x^{2n}}{\sqrt{n+1}} converge? Justify your answer.

    Now, I reckon the sequence converges for -1>x<1, because anything else looks like it will just diverge... but how do I go about "justifying" this?
    Let S_n = \frac{x^{2n}}{\sqrt{n+1}}

    You have to find the radius (and hence the interval) of convergence for this power series.

    You can use the ratio test. Start by finding \mathop {\lim }\limits_{n \to \infty } | \frac{S_{n+1}}{S_n}|= \frac{|x|^{2n+1}}{\sqrt{n+2}} / \frac{|x|^{2n}}{\sqrt{n+1}}
    Last edited by Roam; December 6th 2009 at 12:34 PM.
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  3. #3
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    root test also works here.
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