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Math Help - discuss the convergence/divergence of (annoying function)

  1. #1
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    discuss the convergence/divergence of (annoying function)

    i simply cant get the the limit comparison test to work on this following function,

    Summation of (from 2 to infinity)[ln(k)^(5/2)/k^(15/8)]

    Please someone help!
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  2. #2
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    \forall x > 0,\,\ln (x) < x \Rightarrow \quad \ln \left( {k^{\frac{5}{4}} } \right) < k^{\frac{5}{4}}  \Rightarrow \quad \frac{5}{4}\ln (k) < k^{\frac{5}{4}}

    \frac{{\left[ {\ln (k)} \right]^{\frac{5}{2}} }}{{k^{\frac{{15}}{8}} }} < \frac{{\left( {\frac{4}{5}} \right)^{\frac{2}{5}} k^{\frac{1}{2}} }}{{k^{\frac{{15}}{8}} }} = \frac{{\left( {\frac{4}{5}} \right)^{\frac{2}{5}} }}{{k^{\frac{{11}}{8}} }}<br />
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  3. #3
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    so you're saying that the series converges?? but your b_n is greater than a_n and b_n is convergent by p series equal to 11/8...


    how did u solve it, plz explain more
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  4. #4
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    Quote Originally Posted by ramzouzy View Post
    so you're saying that the series converges?? but your b_n is greater than a_n and b_n is convergent by p series equal to 11/8...
    Number one: I did not use limit comparison.
    I used simple comparison which you can adapt to limit comparison.
    I am no sure that you follow the simple arithmetic involved here.
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  5. #5
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    i am still frustrated... i didnt see how you got from step 3 to step 4 and how did u get <br />
\frac{{\left[ {\ln (k)} \right]^{\frac{5}{2}} }}{{k^{\frac{{15}}{8}} }} < \frac{{\left( {\frac{4}{5}} \right)^{\frac{2}{5}} k^{\frac{1}{2}} }}{{k^{\frac{{15}}{8}} }} = \frac{{\left( {\frac{4}{5}} \right)^{\frac{2}{5}} }}{{k^{\frac{{11}}{8}} }}<br />
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  6. #6
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    O.K.
    Start over. Consider this: \left[ {\ln (k)} \right]^{\frac{5}{2}}  = \left[ {10\ln \left( {k^{\frac{1}{{10}}} } \right)} \right]^{\frac{5}{2}}  < \left( {10} \right)^{\frac{5}{2}} \left( {k^{\frac{1}{{10}}} } \right)^{\frac{5}{2}}  = \left( {10} \right)^{\frac{5}{2}} \left( {k^{\frac{1}{4}} } \right)
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