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Math Help - Convergence Proof

  1. #1
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    Convergence Proof

    if Sn is the summation of partial sums from 1 to infinity of (log n)^-logn, prove whether Sn converges or diverges.

    It gives a hint of using the 2^m test which states that if Am is a nonincreasing function with limit 0 then the Summation of An from 1 to infinityconverges if and only if the series of the summation of partial sums 2^m*A(2^m) from m=0 to m=infinity
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Snooks02 View Post
    if Sn is the summation of partial sums from 1 to infinity of (log n)^-logn, prove whether Sn converges or diverges.

    It gives a hint of using the 2^m test which states that if Am is a nonincreasing function with limit 0 then the Summation of An from 1 to infinityconverges if and only if the series of the summation of partial sums 2^m*A(2^m) from m=0 to m=infinity
    Yeah, just use Cauchy's Condensation test. Since \frac{1}{\ln(n)^{\ln(n)}} is positive and nonincreasing we may apply Cauchy's Condenstaion test which states that \sum_{a_n} converges iff \sum{2^na_{2^n}} does. And in this case we have that \sum{2^na_{2^n}}=\sum\frac{2^n}{(n\ln(2))^{n\ln(2)  }}

    Now you can use whatever tes you want, Ratio, Root, comparison, whatever to determine that that converges.
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