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Math Help - Series C/D ?

  1. #1
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    Series C/D ?

    hi

    am stuck here =S

    sigma(2-->infinity) 1 / (ln n)^9
    i got headache =(
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  2. #2
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    Please any help for this ??
    I belive it can be solved
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  3. #3
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    Quote Originally Posted by TWiX View Post
    hi

    am stuck here =S

    sigma(2-->infinity) 1 / (ln n)^9
    i got headache =(

    Google "Cauchy's Condensation Test": your series diverges.

    Tonio
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  4. #4
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    Quote Originally Posted by tonio View Post
    Google "Cauchy's Condensation Test": your series diverges.

    Tonio
    Thanks, but is it impossible to solve it using the standard tests ?
    i saw it in an incomplete exam in my college =S
    and cauchy's test is not covered in our material..


    TWiX
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  5. #5
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    Quote Originally Posted by TWiX View Post
    Thanks, but is it impossible to solve it using the standard tests ?
    i saw it in an incomplete exam in my college =S
    and cauchy's test is not covered in our material..


    TWiX

    Not as far as I can tell right now, and the condensation test is pretty standard.

    Tonio
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  6. #6
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    Quote Originally Posted by TWiX View Post
    is it impossible to solve it using the standard tests ?
    Use basic comparsion.
    \begin{gathered}<br />
  \ln \left( {\sqrt[9]{n}} \right) \leqslant \sqrt[9]{n} \hfill \\<br />
  \ln (n)^9  \leqslant 9^9 n \hfill \\<br />
  \frac{1}<br />
{{\ln (n)^9 }} \geqslant \frac{{9^{ - 9} }}<br />
{n} \hfill \\ <br />
\end{gathered}
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  7. #7
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    Quote Originally Posted by Plato View Post
    Use basic comparsion.
    \begin{gathered}<br />
\ln \left( {\sqrt[9]{n}} \right) \leqslant \sqrt[9]{n} \hfill \\<br />
\ln (n)^9 \leqslant 9^9 n \hfill \\<br />
\frac{1}<br />
{{\ln (n)^9 }} \geqslant \frac{{9^{ - 9} }}<br />
{n} \hfill \\ <br />
\end{gathered}
    The power is on the "ln n" not just "n"
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  8. #8
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    Quote Originally Posted by TWiX View Post
    The power is on the "ln n" not just "n"
    That is exactly what I wrote: [ln(n)]^9=ln(n)^9=ln^9(n)
    That is perfectly standard notation.

    If it were just on the n it would be ln(n^9) which not what I wrote.

    BTW: If you use a computer algebra system you must use ln(n)^9
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  9. #9
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    Quote Originally Posted by Plato View Post
    That is exactly what I wrote: [ln(n)]^9=ln(n)^9=ln^9(n)
    That is perfectly standard notation.

    If it were just on the n it would be ln(n^9) which not what I wrote.
    idk this :S
    Thank you anyway ^_^
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  10. #10
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    There is another solution
    Limit comparison test with b_n=\frac{1}{n^{\frac{9}{11}}}
    The limit =\infty
    and \sum b_n diverges.
    Hence, Our series diverges.
    It works, Right?
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  11. #11
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by TWiX View Post
    hi

    am stuck here =S

    sigma(2-->infinity) 1 / (ln n)^9
    i got headache =(
    Quote Originally Posted by tonio View Post
    Google "Cauchy's Condensation Test": your series diverges.

    Tonio
    Quote Originally Posted by Plato View Post
    Use basic comparsion.
    \begin{gathered}<br />
  \ln \left( {\sqrt[9]{n}} \right) \leqslant \sqrt[9]{n} \hfill \\<br />
  \ln (n)^9  \leqslant 9^9 n \hfill \\<br />
  \frac{1}<br />
{{\ln (n)^9 }} \geqslant \frac{{9^{ - 9} }}<br />
{n} \hfill \\ <br />
\end{gathered}
    Quote Originally Posted by TWiX View Post
    The power is on the "ln n" not just "n"
    http://www.mathhelpforum.com/math-he...divergent.html
    Did you think you'd get a different answer if you asked twice?
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  12. #12
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    Quote Originally Posted by Drexel28 View Post
    http://www.mathhelpforum.com/math-he...divergent.html
    Did you think you'd get a different answer if you asked twice?
    Who told you am searching for different answer here ?
    and where is the problem If I search for different answers ?
    And in my last post here, I wrote another solution and I asked to check whether its correct or wrong, and you ignore that.
    Yes, You answered in another thread and I thanked you.
    But where is the problem If I'm searching for different answers ?
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  13. #13
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by TWiX View Post
    Who told you am searching for different answer here ?
    and where is the problem If I search for different answers ?
    And in my last post here, I wrote another solution and I asked to check whether its correct or wrong, and you ignore that.
    Yes, You answered in another thread and I thanked you.
    But where is the problem If I'm searching for different answers ?
    Haha, I hope my last post didn't sound annoyed. I forgot to put "". What was your question?
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  14. #14
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    Here is it


    Quote Originally Posted by TWiX View Post
    There is another solution
    Limit comparison test with b_n=\frac{1}{n^{\frac{9}{11}}}
    The limit =\infty
    and \sum b_n diverges.
    Hence, Our series diverges.
    It works, Right?
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  15. #15
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by TWiX View Post
    Here is it
    Write explicitly what limit test you used.
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