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Math Help - Radii of convergence of a power series

  1. #1
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    Radii of convergence of a power series

    How do I compute the radii of convergence of the power series Σanx^n

    (from n=0 to infinity)

    with coefficients an=n and an=1/n
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  2. #2
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    Quote Originally Posted by TexasGirl
    How do I compute the radii of convergence of the power series Σanx^n

    (from n=0 to infinity)

    with coefficients an=n and an=1/n
    Use the ratio test for the power series. With a_k=k thus, \lim_{k\rightarrow \infty}|\frac{k+1}{k}|=1 Thus, the radius of convergence is the the reciprocal of that thus 1/1=1. However, the interval of convergence is -1<x<1. For x=-1,1 this power series diverges.

    Now for the second problem, a_k=1/k, k>1 use the ratio test for power series again to get \lim_{k\rightarrow \infty}|\frac{k}{k+1}|=1 Thus, the radius of converges is 1, thus for -1<x<1 converges absolutely. Checking the endpoint (because the ratio test is inconclusive for when its limit is one), we have
    \sum^\infty_{k=1} (-1)^k 1/k but this is the alternating-harmonic series thus it converges. For the second possibility we have that \sum^\infty_{k=1} 1/k but this is the harmonic series thus it diverges. Thus, the interval of convergence for the second power series is -1\leq x<1
    Last edited by ThePerfectHacker; January 8th 2006 at 11:38 AM.
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  3. #3
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    One More Radius of Convergence Question...

    Building from the same power series, if I take q, a nonzero element of C, and put it into the series so that I have the sum of an(q^n)(x^n), can I still use the ratio test in order to find the radius of convergence? Is an the coefficient?
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  4. #4
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    Quote Originally Posted by TexasGirl
    Building from the same power series, if I take q, a nonzero element of C, and put it into the series so that I have the sum of an(q^n)(x^n), can I still use the ratio test in order to find the radius of convergence? Is an the coefficient?
    I do not understand what you are asking?
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  5. #5
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    Here is the question I have to answer:

    Assume that the power series ∑an(x^n), where n=0 to infinity, has radius of convergence p, where p is a nonnegative real number or stands for the symbol ∞. Let q be an element of C, q≠0. Compute the radius of convergence of ∑an(q^n)(x^n) (n=0 to infinity).
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  6. #6
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    As I understand it, its radius of convergence would be p/q Explanation:
    If \sum^{\infty}_{k=1}a_kx^k has radius of convergence of p then by the ratio test for power series \lim_{k\rightarrow \infty}|\frac{a_{k+1}}{a_k}|=1/p because as I said it is the reciprocal of the limit. Thus, the new infinite series given by \sum^{\infty}_{k=1}a_kq^kx^k then by the ratio test \lim_{k\rightarrow \infty}|\frac{a_{k+1}q^{k+1}}{a_kq^k}|=\frac{q}{p} thus, the radius of convergent is the reciprocal thus \frac{p}{q}
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  7. #7
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    thanks again...

    much appreciated...
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  8. #8
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    You are welcome.
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