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Math Help - Radius and interval of convergence?

  1. #1
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    Radius and interval of convergence?

    I am trying to find the series' radius and interval of convergence and for what values of x the series converge b) absolutely, c) conditionally?

    1) n = 1 to infinite, n^r*x^n, r real.
    2) n = 1 to infinite, (3^n(x-1)^n)/n^3
    3) n = 1 to infinite, ((n!)^2 * x^n)/(2n)!

    So I know I need to apply the ratio test, and that left me with ((n+1)^r*x)/n^r for the first one, (3(x-1)*n^3)/(n+1)^3 for the second, and ((n+1)^2*x)/2(n+1) for the third-- assuming I did this correctly. I just don't understand what to do next. Any help would be very appreciated!
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  2. #2
    Junior Member
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    Nov 2009
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    Those ratios look correct. Now, take the limit as  n \rightarrow \infty of the absolute value of the ratios you've computed. The series converges absolutely if that limit is less than 1 and diverges if that limit is greater than 1. If the limit is 1 exactly, the ratio test is inconclusive and you will have to use some other test to determine convergence.

    For the second series, for example, the limit as  n \rightarrow \infty of the absolute value of the ratio is 3|x-1|, which is less than 1 if 2/3 < x < 4/3, equals 1 if x = 2/3 or x = 4/3, and is greater than 1 everywhere else. This tells you that the series converges absolutely if 2/3 < x < 4/3, diverges if x < 2/3 or x > 4/3, but tells you nothing of convergence if x = 2/3 or x = 4/3. That is, for these two values of x you must test for convergence using other means. When x = 2/3, you get  \sum_{n=1}^{\infty} \frac{(-1)^n}{n^3} and when x = 4/3 you get  \sum_{n = 1}^{\infty} \frac{1}{n^3} , both of which converge absolutely. So the series converges absolutely for  2/3 \le x \le 4/3 and diverges everywhere else.
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