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Math Help - Alternate series converges

  1. #1
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    Alternate series converges

    Prove that (1-2) - (1-2^(1/2)) + (1 - 2^(1/3)) - (1 - 2^(1/4)) + ... converges.
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by tarheelborn View Post
    Prove that (1-2) - (1-2^(1/2)) + (1 - 2^(1/3)) - (1 - 2^(1/4)) + ... converges.
    Just show  \sqrt[n]{2}\to1 monotonically, then just apply the alternating series test.
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    Ah, so since (1-2) - (1- 2^(1/2))+... = (-1)(2-1) + (1)(2^(1/2) -1)-... = \sum (-1)^n(2^(1/n)-1). Lim 2^(1/n) = 1 ==> Lim 2^(1/n)-1 = 0. Also, 2^(1/n)>=2^1/(n+1)>=2^(1/n+2)... so {2^(1/n)} is nonincreasing. Thus this series is convergent by alternate series test. That will work, right? Thanks!
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by tarheelborn View Post
    Ah, so since (1-2) - (1- 2^(1/2))+... = (-1)(2-1) + (1)(2^(1/2) -1)-... = \sum (-1)^n(2^(1/n)-1). Lim 2^(1/n) = 1 ==> Lim 2^(1/n)-1 = 0. Also, 2^(1/n)>=2^1/(n+1)>=2^(1/n+2)... so {2^(1/n)} is nonincreasing. Thus this series is convergent by alternate series test. That will work, right? Thanks!
    Correct. You might want to say why  2^{\frac1n}>2^{\frac{1}{n+1}} .
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