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Math Help - convergence again..... ><

  1. #1
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    convergence again..... ><

    sum n=1 oo [(-1)^n / n^2/3 + 1 / n^3/2]

    so the second part is easy. p series 3/2>1 converges

    but i cant get the first part? im too familiar with the alternating series test

    ((-1)^n) 1 / n^2/3

    do you just try to find the limit of that by doing

    lim n->oo 1 / n^2/3 = 1/oo = 0?
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  2. #2
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    Quote Originally Posted by jeph View Post
    sum n=1 oo [(-1)^n / n^2/3 + 1 / n^3/2]

    so the second part is easy. p series 3/2>1 converges

    but i cant get the first part? im too familiar with the alternating series test

    ((-1)^n) 1 / n^2/3

    do you just try to find the limit of that by doing

    lim n->oo 1 / n^2/3 = 1/oo = 0?
    You should try to apply the following theorem:
    Alternating series test: A series of the form ∑[ (−1)^n] an (with an ≥ 0) is called alternating. Such a series converges if the sequence an is monotone decreasing and converges to 0. The converse is in general not true.

    You noticed that bn= 1/n^(2/3) converges to 0.
    All you have to do is to show that is also decreasing.
    Use a ratio test:bn+1/bn=n^2/3/(n+1)^2/3=(n/n+1)^2/3<1, so decreasing.
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  3. #3
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    how do you show (n/n+1)^2/3<1
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  4. #4
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    Since n goes from 1 to infinity, is a positive
    So, n<n+1,and n^2/3<(n+1)^2/3
    Therefore, [n/(n+1)]^2/3<1
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