Discuss the convergence of the series $sum_{n=2}^infty(ln{n})^{-ln(ln{n})}$












1












$begingroup$


Discuss the convergence or divergence of the series where the general term $x_n$ is given by



$$x_n=(ln{n})^{-ln(ln{n})},,,nge 2.$$






calculus real-analysis sequences-and-series limits convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    Cauchy's Condensation Test gives you a clearly divergent series (its general term doesn't even converge to zero...)
    $endgroup$
    – DonAntonio
    Mar 16 '14 at 13:20










  • $begingroup$
    I have not learnt Cauchy Condensation Test. Can you tell me how it works? Thanks
    $endgroup$
    – anonymous
    Mar 16 '14 at 13:26










  • $begingroup$
    Google it, @anonymous....:)
    $endgroup$
    – DonAntonio
    Mar 16 '14 at 13:27










  • $begingroup$
    @DonAntonio: Are you saying $x_n not to 0$ as $n to infty$?
    $endgroup$
    – Henry
    Mar 16 '14 at 13:29






  • 2




    $begingroup$
    @Henry No, he's saying $2^kcdot x_{2^k} notto 0$. (Or maybe replace the $2$ with a different base, the condensation test isn't tied to $2$.)
    $endgroup$
    – Daniel Fischer♦
    Mar 16 '14 at 13:32
















1












$begingroup$


Discuss the convergence or divergence of the series where the general term $x_n$ is given by



$$x_n=(ln{n})^{-ln(ln{n})},,,nge 2.$$






calculus real-analysis sequences-and-series limits convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    Cauchy's Condensation Test gives you a clearly divergent series (its general term doesn't even converge to zero...)
    $endgroup$
    – DonAntonio
    Mar 16 '14 at 13:20










  • $begingroup$
    I have not learnt Cauchy Condensation Test. Can you tell me how it works? Thanks
    $endgroup$
    – anonymous
    Mar 16 '14 at 13:26










  • $begingroup$
    Google it, @anonymous....:)
    $endgroup$
    – DonAntonio
    Mar 16 '14 at 13:27










  • $begingroup$
    @DonAntonio: Are you saying $x_n not to 0$ as $n to infty$?
    $endgroup$
    – Henry
    Mar 16 '14 at 13:29






  • 2




    $begingroup$
    @Henry No, he's saying $2^kcdot x_{2^k} notto 0$. (Or maybe replace the $2$ with a different base, the condensation test isn't tied to $2$.)
    $endgroup$
    – Daniel Fischer♦
    Mar 16 '14 at 13:32














1












1








1


1



$begingroup$


Discuss the convergence or divergence of the series where the general term $x_n$ is given by



$$x_n=(ln{n})^{-ln(ln{n})},,,nge 2.$$






calculus real-analysis sequences-and-series limits convergence







share|cite|improve this question











$endgroup$




Discuss the convergence or divergence of the series where the general term $x_n$ is given by



$$x_n=(ln{n})^{-ln(ln{n})},,,nge 2.$$






calculus real-analysis sequences-and-series limits convergence




calculus real-analysis sequences-and-series limits convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 23 '14 at 14:55









Yiorgos S. Smyrlis

62.9k1384163




62.9k1384163










asked Mar 16 '14 at 13:15









anonymousanonymous

211




211












  • $begingroup$
    Cauchy's Condensation Test gives you a clearly divergent series (its general term doesn't even converge to zero...)
    $endgroup$
    – DonAntonio
    Mar 16 '14 at 13:20










  • $begingroup$
    I have not learnt Cauchy Condensation Test. Can you tell me how it works? Thanks
    $endgroup$
    – anonymous
    Mar 16 '14 at 13:26










  • $begingroup$
    Google it, @anonymous....:)
    $endgroup$
    – DonAntonio
    Mar 16 '14 at 13:27










  • $begingroup$
    @DonAntonio: Are you saying $x_n not to 0$ as $n to infty$?
    $endgroup$
    – Henry
    Mar 16 '14 at 13:29






  • 2




    $begingroup$
    @Henry No, he's saying $2^kcdot x_{2^k} notto 0$. (Or maybe replace the $2$ with a different base, the condensation test isn't tied to $2$.)
    $endgroup$
    – Daniel Fischer♦
    Mar 16 '14 at 13:32


















  • $begingroup$
    Cauchy's Condensation Test gives you a clearly divergent series (its general term doesn't even converge to zero...)
    $endgroup$
    – DonAntonio
    Mar 16 '14 at 13:20










  • $begingroup$
    I have not learnt Cauchy Condensation Test. Can you tell me how it works? Thanks
    $endgroup$
    – anonymous
    Mar 16 '14 at 13:26










  • $begingroup$
    Google it, @anonymous....:)
    $endgroup$
    – DonAntonio
    Mar 16 '14 at 13:27










  • $begingroup$
    @DonAntonio: Are you saying $x_n not to 0$ as $n to infty$?
    $endgroup$
    – Henry
    Mar 16 '14 at 13:29






  • 2




    $begingroup$
    @Henry No, he's saying $2^kcdot x_{2^k} notto 0$. (Or maybe replace the $2$ with a different base, the condensation test isn't tied to $2$.)
    $endgroup$
    – Daniel Fischer♦
    Mar 16 '14 at 13:32
















$begingroup$
Cauchy's Condensation Test gives you a clearly divergent series (its general term doesn't even converge to zero...)
$endgroup$
– DonAntonio
Mar 16 '14 at 13:20




$begingroup$
Cauchy's Condensation Test gives you a clearly divergent series (its general term doesn't even converge to zero...)
$endgroup$
– DonAntonio
Mar 16 '14 at 13:20












$begingroup$
I have not learnt Cauchy Condensation Test. Can you tell me how it works? Thanks
$endgroup$
– anonymous
Mar 16 '14 at 13:26




$begingroup$
I have not learnt Cauchy Condensation Test. Can you tell me how it works? Thanks
$endgroup$
– anonymous
Mar 16 '14 at 13:26












$begingroup$
Google it, @anonymous....:)
$endgroup$
– DonAntonio
Mar 16 '14 at 13:27




$begingroup$
Google it, @anonymous....:)
$endgroup$
– DonAntonio
Mar 16 '14 at 13:27












$begingroup$
@DonAntonio: Are you saying $x_n not to 0$ as $n to infty$?
$endgroup$
– Henry
Mar 16 '14 at 13:29




$begingroup$
@DonAntonio: Are you saying $x_n not to 0$ as $n to infty$?
$endgroup$
– Henry
Mar 16 '14 at 13:29




2




2




$begingroup$
@Henry No, he's saying $2^kcdot x_{2^k} notto 0$. (Or maybe replace the $2$ with a different base, the condensation test isn't tied to $2$.)
$endgroup$
– Daniel Fischer♦
Mar 16 '14 at 13:32




$begingroup$
@Henry No, he's saying $2^kcdot x_{2^k} notto 0$. (Or maybe replace the $2$ with a different base, the condensation test isn't tied to $2$.)
$endgroup$
– Daniel Fischer♦
Mar 16 '14 at 13:32










1 Answer
1






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$begingroup$

We have that
$$
(ln n)^{-ln ln n}=mathrm{e}^{-(ln ln n)^2}>mathrm{e}^{-ln n}=frac{1}{n},
$$
for $n$ sufficiently large since
$$
lim_{ntoinfty}frac{(ln ln n)^2}{ln n}=lim_{Ntoinfty}frac{(ln N)^2}{N}=0.
$$
Hence the series
$$
sum_{n=2}^infty(ln n)^{-ln ln n}
$$
diverges to infinity, by virtue of the comparison test.






share|cite|improve this answer









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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

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    10












    $begingroup$

    We have that
    $$
    (ln n)^{-ln ln n}=mathrm{e}^{-(ln ln n)^2}>mathrm{e}^{-ln n}=frac{1}{n},
    $$
    for $n$ sufficiently large since
    $$
    lim_{ntoinfty}frac{(ln ln n)^2}{ln n}=lim_{Ntoinfty}frac{(ln N)^2}{N}=0.
    $$
    Hence the series
    $$
    sum_{n=2}^infty(ln n)^{-ln ln n}
    $$
    diverges to infinity, by virtue of the comparison test.






    share|cite|improve this answer









    $endgroup$


















      10












      $begingroup$

      We have that
      $$
      (ln n)^{-ln ln n}=mathrm{e}^{-(ln ln n)^2}>mathrm{e}^{-ln n}=frac{1}{n},
      $$
      for $n$ sufficiently large since
      $$
      lim_{ntoinfty}frac{(ln ln n)^2}{ln n}=lim_{Ntoinfty}frac{(ln N)^2}{N}=0.
      $$
      Hence the series
      $$
      sum_{n=2}^infty(ln n)^{-ln ln n}
      $$
      diverges to infinity, by virtue of the comparison test.






      share|cite|improve this answer









      $endgroup$
















        10












        10








        10





        $begingroup$

        We have that
        $$
        (ln n)^{-ln ln n}=mathrm{e}^{-(ln ln n)^2}>mathrm{e}^{-ln n}=frac{1}{n},
        $$
        for $n$ sufficiently large since
        $$
        lim_{ntoinfty}frac{(ln ln n)^2}{ln n}=lim_{Ntoinfty}frac{(ln N)^2}{N}=0.
        $$
        Hence the series
        $$
        sum_{n=2}^infty(ln n)^{-ln ln n}
        $$
        diverges to infinity, by virtue of the comparison test.






        share|cite|improve this answer









        $endgroup$



        We have that
        $$
        (ln n)^{-ln ln n}=mathrm{e}^{-(ln ln n)^2}>mathrm{e}^{-ln n}=frac{1}{n},
        $$
        for $n$ sufficiently large since
        $$
        lim_{ntoinfty}frac{(ln ln n)^2}{ln n}=lim_{Ntoinfty}frac{(ln N)^2}{N}=0.
        $$
        Hence the series
        $$
        sum_{n=2}^infty(ln n)^{-ln ln n}
        $$
        diverges to infinity, by virtue of the comparison test.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 16 '14 at 13:29









        Yiorgos S. SmyrlisYiorgos S. Smyrlis

        62.9k1384163




        62.9k1384163






























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