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Thread: Series

  1. #1
    Feb 2006


    Please help...

    1.) find the series' radius and interval of convergence. for what values of x does the series converge absolutely and conditionally?

    Σ [(3^n)(x^n)] / (n!)

    2.) find the sum of the series by expressing 1/ (1-x) as a geometric series

    Σ (n^2)/(2^n)
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  2. #2
    Senior Member
    Jun 2005
    A handy test in this area is the n-th root test: given a series \sum_n a_n, for which the sequence a_n^{1/n} tends to a limit \ell, the series is convergent if \ell < 1 and divergent if \ell > 1. (In the case \ell=1, the test gives no information.)

    Applied to your series (1), we see that (3^n x^n / n!)^{1/n} = 3x/(n!)^{1/n} and it isn't too hard to show that (n!)^{1/n} \rightarrow \infty. Hence the limit \ell = 0 and the series is convergent for any value of x.

    For your series (2) a useful trick is to regard it as a special value of a function f(x) = \sum_n n^2 x^n, evaluated at x=1/2. We start with 1/(1-x) = \sum_n x^n and differentiate to get 1/(1-x)^2 = \sum_n n x^{n-1}. Multiply by x and differentiate again to find f.
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