A test for convergence involving logarithms












1












$begingroup$


$$sum_{n=2}^infty frac{n^{log n}}{(log n)^n}$$



Tried to use the inequality of $ln x< x-1$ and apply the comparison theorem but no success.






calculus sequences-and-series







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  • $begingroup$
    As tip for future posts, you can enclose exponents in braces { } to capture the entire argument. That is, $n^{log n}$ produces $n^{log n}$.
    $endgroup$
    – Clayton
    Jan 30 at 1:54
















1












$begingroup$


$$sum_{n=2}^infty frac{n^{log n}}{(log n)^n}$$



Tried to use the inequality of $ln x< x-1$ and apply the comparison theorem but no success.






calculus sequences-and-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    As tip for future posts, you can enclose exponents in braces { } to capture the entire argument. That is, $n^{log n}$ produces $n^{log n}$.
    $endgroup$
    – Clayton
    Jan 30 at 1:54














1












1








1





$begingroup$


$$sum_{n=2}^infty frac{n^{log n}}{(log n)^n}$$



Tried to use the inequality of $ln x< x-1$ and apply the comparison theorem but no success.






calculus sequences-and-series







share|cite|improve this question











$endgroup$




$$sum_{n=2}^infty frac{n^{log n}}{(log n)^n}$$



Tried to use the inequality of $ln x< x-1$ and apply the comparison theorem but no success.






calculus sequences-and-series




calculus sequences-and-series






share|cite|improve this question















share|cite|improve this question













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edited Jan 30 at 3:03









David G. Stork

11.6k41534




11.6k41534










asked Jan 30 at 1:52









Nick KaklNick Kakl

176




176












  • $begingroup$
    As tip for future posts, you can enclose exponents in braces { } to capture the entire argument. That is, $n^{log n}$ produces $n^{log n}$.
    $endgroup$
    – Clayton
    Jan 30 at 1:54


















  • $begingroup$
    As tip for future posts, you can enclose exponents in braces { } to capture the entire argument. That is, $n^{log n}$ produces $n^{log n}$.
    $endgroup$
    – Clayton
    Jan 30 at 1:54
















$begingroup$
As tip for future posts, you can enclose exponents in braces { } to capture the entire argument. That is, $n^{log n}$ produces $n^{log n}$.
$endgroup$
– Clayton
Jan 30 at 1:54




$begingroup$
As tip for future posts, you can enclose exponents in braces { } to capture the entire argument. That is, $n^{log n}$ produces $n^{log n}$.
$endgroup$
– Clayton
Jan 30 at 1:54










3 Answers
3






active

oldest

votes


















1












$begingroup$

We consider
$$lim_{nto infty}sqrt[n]{frac{n^{log n}}{(log n)^n}}
=lim_{nto infty}frac{n^{frac{log n}{n}}}{log n}
=lim_{nto infty}frac{e^{frac{(log n)^2}{n}}}{log n}=0.$$
By Cauchy test, this series is convergent.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thnx for the answer, you got the limit with DLH?
    $endgroup$
    – Nick Kakl
    Jan 30 at 2:59










  • $begingroup$
    Yse, it is easy to get $frac{(log n)^2}{n}to 0$ as $nto infty$.
    $endgroup$
    – Riemann
    Jan 30 at 3:03





















1












$begingroup$

Let's upper-bound each term with a term whose sum converges. Note that, for $n$ large enough, $log log n geq 1$, therefore:
$$frac{n^{log n}}{(log n)^n} = frac{e^{log^2n}}{e^{nloglog n}}leq frac{e^{log^2n}}{e^{n}}=e^{-frac n 2}e^{-frac n 2+log^2n}$$
and because for $n$ large enough, $left(-frac n 2+log^2nright)leq 0$ (left as an exercise),
we must have $$frac{n^{log n}}{(log n)^n} leq e^{-frac n 2}$$
Conclusion: The sum converges






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ty for the answer :)
    $endgroup$
    – Nick Kakl
    Jan 30 at 2:59



















0












$begingroup$

Too long for a comment.



The ratio test is interesting to work
$$a_n=frac{n^{log n}}{(log n)^n}implies log(a_n)=log ^2(n)-n log (n)$$ Now, using Taylor for infinitely large values of $n$,
$$log(a_{n+1})-log(a_n)=-1-log left({n}right)+frac{2 log
left({n}right)-frac{1}{2}}{n}+Oleft(frac{1}{n^2}right)$$ and, continuing with Taylor expansions,
$$frac{a_{n+1} } {a_n }=e^{log(a_{n+1})-log(a_n) }=frac1 {e n}+frac{2 log
left({n}right)-frac{1}{2}}{en^2}+Oleft(frac{1}{n^3}right)$$






share|cite|improve this answer









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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    1












    $begingroup$

    We consider
    $$lim_{nto infty}sqrt[n]{frac{n^{log n}}{(log n)^n}}
    =lim_{nto infty}frac{n^{frac{log n}{n}}}{log n}
    =lim_{nto infty}frac{e^{frac{(log n)^2}{n}}}{log n}=0.$$
    By Cauchy test, this series is convergent.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thnx for the answer, you got the limit with DLH?
      $endgroup$
      – Nick Kakl
      Jan 30 at 2:59










    • $begingroup$
      Yse, it is easy to get $frac{(log n)^2}{n}to 0$ as $nto infty$.
      $endgroup$
      – Riemann
      Jan 30 at 3:03


















    1












    $begingroup$

    We consider
    $$lim_{nto infty}sqrt[n]{frac{n^{log n}}{(log n)^n}}
    =lim_{nto infty}frac{n^{frac{log n}{n}}}{log n}
    =lim_{nto infty}frac{e^{frac{(log n)^2}{n}}}{log n}=0.$$
    By Cauchy test, this series is convergent.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thnx for the answer, you got the limit with DLH?
      $endgroup$
      – Nick Kakl
      Jan 30 at 2:59










    • $begingroup$
      Yse, it is easy to get $frac{(log n)^2}{n}to 0$ as $nto infty$.
      $endgroup$
      – Riemann
      Jan 30 at 3:03
















    1












    1








    1





    $begingroup$

    We consider
    $$lim_{nto infty}sqrt[n]{frac{n^{log n}}{(log n)^n}}
    =lim_{nto infty}frac{n^{frac{log n}{n}}}{log n}
    =lim_{nto infty}frac{e^{frac{(log n)^2}{n}}}{log n}=0.$$
    By Cauchy test, this series is convergent.






    share|cite|improve this answer









    $endgroup$



    We consider
    $$lim_{nto infty}sqrt[n]{frac{n^{log n}}{(log n)^n}}
    =lim_{nto infty}frac{n^{frac{log n}{n}}}{log n}
    =lim_{nto infty}frac{e^{frac{(log n)^2}{n}}}{log n}=0.$$
    By Cauchy test, this series is convergent.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 30 at 2:54









    RiemannRiemann

    3,5251422




    3,5251422












    • $begingroup$
      Thnx for the answer, you got the limit with DLH?
      $endgroup$
      – Nick Kakl
      Jan 30 at 2:59










    • $begingroup$
      Yse, it is easy to get $frac{(log n)^2}{n}to 0$ as $nto infty$.
      $endgroup$
      – Riemann
      Jan 30 at 3:03




















    • $begingroup$
      Thnx for the answer, you got the limit with DLH?
      $endgroup$
      – Nick Kakl
      Jan 30 at 2:59










    • $begingroup$
      Yse, it is easy to get $frac{(log n)^2}{n}to 0$ as $nto infty$.
      $endgroup$
      – Riemann
      Jan 30 at 3:03


















    $begingroup$
    Thnx for the answer, you got the limit with DLH?
    $endgroup$
    – Nick Kakl
    Jan 30 at 2:59




    $begingroup$
    Thnx for the answer, you got the limit with DLH?
    $endgroup$
    – Nick Kakl
    Jan 30 at 2:59












    $begingroup$
    Yse, it is easy to get $frac{(log n)^2}{n}to 0$ as $nto infty$.
    $endgroup$
    – Riemann
    Jan 30 at 3:03






    $begingroup$
    Yse, it is easy to get $frac{(log n)^2}{n}to 0$ as $nto infty$.
    $endgroup$
    – Riemann
    Jan 30 at 3:03













    1












    $begingroup$

    Let's upper-bound each term with a term whose sum converges. Note that, for $n$ large enough, $log log n geq 1$, therefore:
    $$frac{n^{log n}}{(log n)^n} = frac{e^{log^2n}}{e^{nloglog n}}leq frac{e^{log^2n}}{e^{n}}=e^{-frac n 2}e^{-frac n 2+log^2n}$$
    and because for $n$ large enough, $left(-frac n 2+log^2nright)leq 0$ (left as an exercise),
    we must have $$frac{n^{log n}}{(log n)^n} leq e^{-frac n 2}$$
    Conclusion: The sum converges






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Ty for the answer :)
      $endgroup$
      – Nick Kakl
      Jan 30 at 2:59
















    1












    $begingroup$

    Let's upper-bound each term with a term whose sum converges. Note that, for $n$ large enough, $log log n geq 1$, therefore:
    $$frac{n^{log n}}{(log n)^n} = frac{e^{log^2n}}{e^{nloglog n}}leq frac{e^{log^2n}}{e^{n}}=e^{-frac n 2}e^{-frac n 2+log^2n}$$
    and because for $n$ large enough, $left(-frac n 2+log^2nright)leq 0$ (left as an exercise),
    we must have $$frac{n^{log n}}{(log n)^n} leq e^{-frac n 2}$$
    Conclusion: The sum converges






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Ty for the answer :)
      $endgroup$
      – Nick Kakl
      Jan 30 at 2:59














    1












    1








    1





    $begingroup$

    Let's upper-bound each term with a term whose sum converges. Note that, for $n$ large enough, $log log n geq 1$, therefore:
    $$frac{n^{log n}}{(log n)^n} = frac{e^{log^2n}}{e^{nloglog n}}leq frac{e^{log^2n}}{e^{n}}=e^{-frac n 2}e^{-frac n 2+log^2n}$$
    and because for $n$ large enough, $left(-frac n 2+log^2nright)leq 0$ (left as an exercise),
    we must have $$frac{n^{log n}}{(log n)^n} leq e^{-frac n 2}$$
    Conclusion: The sum converges






    share|cite|improve this answer









    $endgroup$



    Let's upper-bound each term with a term whose sum converges. Note that, for $n$ large enough, $log log n geq 1$, therefore:
    $$frac{n^{log n}}{(log n)^n} = frac{e^{log^2n}}{e^{nloglog n}}leq frac{e^{log^2n}}{e^{n}}=e^{-frac n 2}e^{-frac n 2+log^2n}$$
    and because for $n$ large enough, $left(-frac n 2+log^2nright)leq 0$ (left as an exercise),
    we must have $$frac{n^{log n}}{(log n)^n} leq e^{-frac n 2}$$
    Conclusion: The sum converges







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 30 at 2:49









    Stefan LafonStefan Lafon

    3,005212




    3,005212












    • $begingroup$
      Ty for the answer :)
      $endgroup$
      – Nick Kakl
      Jan 30 at 2:59


















    • $begingroup$
      Ty for the answer :)
      $endgroup$
      – Nick Kakl
      Jan 30 at 2:59
















    $begingroup$
    Ty for the answer :)
    $endgroup$
    – Nick Kakl
    Jan 30 at 2:59




    $begingroup$
    Ty for the answer :)
    $endgroup$
    – Nick Kakl
    Jan 30 at 2:59











    0












    $begingroup$

    Too long for a comment.



    The ratio test is interesting to work
    $$a_n=frac{n^{log n}}{(log n)^n}implies log(a_n)=log ^2(n)-n log (n)$$ Now, using Taylor for infinitely large values of $n$,
    $$log(a_{n+1})-log(a_n)=-1-log left({n}right)+frac{2 log
    left({n}right)-frac{1}{2}}{n}+Oleft(frac{1}{n^2}right)$$     and, continuing with Taylor expansions,
    $$frac{a_{n+1} } {a_n }=e^{log(a_{n+1})-log(a_n) }=frac1 {e n}+frac{2 log
    left({n}right)-frac{1}{2}}{en^2}+Oleft(frac{1}{n^3}right)$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Too long for a comment.



      The ratio test is interesting to work
      $$a_n=frac{n^{log n}}{(log n)^n}implies log(a_n)=log ^2(n)-n log (n)$$ Now, using Taylor for infinitely large values of $n$,
      $$log(a_{n+1})-log(a_n)=-1-log left({n}right)+frac{2 log
      left({n}right)-frac{1}{2}}{n}+Oleft(frac{1}{n^2}right)$$       and, continuing with Taylor expansions,
      $$frac{a_{n+1} } {a_n }=e^{log(a_{n+1})-log(a_n) }=frac1 {e n}+frac{2 log
      left({n}right)-frac{1}{2}}{en^2}+Oleft(frac{1}{n^3}right)$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Too long for a comment.



        The ratio test is interesting to work
        $$a_n=frac{n^{log n}}{(log n)^n}implies log(a_n)=log ^2(n)-n log (n)$$ Now, using Taylor for infinitely large values of $n$,
        $$log(a_{n+1})-log(a_n)=-1-log left({n}right)+frac{2 log
        left({n}right)-frac{1}{2}}{n}+Oleft(frac{1}{n^2}right)$$         and, continuing with Taylor expansions,
        $$frac{a_{n+1} } {a_n }=e^{log(a_{n+1})-log(a_n) }=frac1 {e n}+frac{2 log
        left({n}right)-frac{1}{2}}{en^2}+Oleft(frac{1}{n^3}right)$$






        share|cite|improve this answer









        $endgroup$



        Too long for a comment.



        The ratio test is interesting to work
        $$a_n=frac{n^{log n}}{(log n)^n}implies log(a_n)=log ^2(n)-n log (n)$$ Now, using Taylor for infinitely large values of $n$,
        $$log(a_{n+1})-log(a_n)=-1-log left({n}right)+frac{2 log
        left({n}right)-frac{1}{2}}{n}+Oleft(frac{1}{n^2}right)$$         and, continuing with Taylor expansions,
        $$frac{a_{n+1} } {a_n }=e^{log(a_{n+1})-log(a_n) }=frac1 {e n}+frac{2 log
        left({n}right)-frac{1}{2}}{en^2}+Oleft(frac{1}{n^3}right)$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 30 at 5:45









        Claude LeiboviciClaude Leibovici

        125k1158136




        125k1158136






























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