The convergence of this series: $sumlimits_{n=2}^infty {1over n^{log n}}$what are some of the methods to do...

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The convergence of this series: $sumlimits_{n=2}^infty {1over n^{log n}}$


what are some of the methods to do divergent/convergent?Determining convergence/divergence with root testConvergence of the series $sum limits_{n=2}^{infty} frac{1}{nlog^s n}$Is $sum_{k=4}^{infty }{k^{log(k)}}/{(log(k))^{k}}$ convergent or divergent?How to prove that if $sum|a_j|<infty$, then $sum|a_j|log(|a_j|^{-1})<infty$Convergence of the series $sumlimits_{n=3}^infty (loglog n)^{-loglog n}$Finding interval of convergence for complicated sumseries convergence with comparison test $frac{1}{nlog(n)^p}$Convergence of series: hintStudying the convergence of the series $sum_{n=1}^inftysinfrac1{n}logleft(1+sinfrac1{n}right)$Convergence of the series $sumlimits_{n=1}^{infty}frac1nlogleft(1+frac1nright)$.Convergence of the series $sum_{n=1}^{infty}frac{(-1)^nsqrt[n]n}{log n}$.













2












$begingroup$


I came across a problem on convergence of series and I did not get into any idea about this -- any help or hints ?
$$sum_{n=2}^infty {1over n^{log n}}$$



What about $n^{log(log n)}$ ?






real-analysis sequences-and-series convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    Making a guess as to what you mean, Baby Rudin 3.28-3.29 would seem helpful.
    $endgroup$
    – ForgotALot
    Mar 7 '16 at 5:59






  • 1




    $begingroup$
    To show divergence if $ale 1$, it is enough (why?) to show divergence when $n=1$. For this use the Integral Test.
    $endgroup$
    – André Nicolas
    Mar 7 '16 at 6:11












  • $begingroup$
    Hello can you make it clear that my question is about n power logn
    $endgroup$
    – Mathslover shah
    Mar 7 '16 at 9:20






  • 3




    $begingroup$
    For some $N$, $n>Nimplieslog(n)>1$. For some other $N$, $n>Nimplieslog(log(n))>1$.
    $endgroup$
    – Yves Daoust
    Mar 7 '16 at 9:24












  • $begingroup$
    Can you edit the problem with n power to the logn please i am not able to do so
    $endgroup$
    – Mathslover shah
    Mar 7 '16 at 9:25
















2












$begingroup$


I came across a problem on convergence of series and I did not get into any idea about this -- any help or hints ?
$$sum_{n=2}^infty {1over n^{log n}}$$



What about $n^{log(log n)}$ ?






real-analysis sequences-and-series convergence







share|cite|improve this question











$endgroup$












  • $begingroup$
    Making a guess as to what you mean, Baby Rudin 3.28-3.29 would seem helpful.
    $endgroup$
    – ForgotALot
    Mar 7 '16 at 5:59






  • 1




    $begingroup$
    To show divergence if $ale 1$, it is enough (why?) to show divergence when $n=1$. For this use the Integral Test.
    $endgroup$
    – André Nicolas
    Mar 7 '16 at 6:11












  • $begingroup$
    Hello can you make it clear that my question is about n power logn
    $endgroup$
    – Mathslover shah
    Mar 7 '16 at 9:20






  • 3




    $begingroup$
    For some $N$, $n>Nimplieslog(n)>1$. For some other $N$, $n>Nimplieslog(log(n))>1$.
    $endgroup$
    – Yves Daoust
    Mar 7 '16 at 9:24












  • $begingroup$
    Can you edit the problem with n power to the logn please i am not able to do so
    $endgroup$
    – Mathslover shah
    Mar 7 '16 at 9:25














2












2








2





$begingroup$


I came across a problem on convergence of series and I did not get into any idea about this -- any help or hints ?
$$sum_{n=2}^infty {1over n^{log n}}$$



What about $n^{log(log n)}$ ?






real-analysis sequences-and-series convergence







share|cite|improve this question











$endgroup$




I came across a problem on convergence of series and I did not get into any idea about this -- any help or hints ?
$$sum_{n=2}^infty {1over n^{log n}}$$



What about $n^{log(log n)}$ ?






real-analysis sequences-and-series convergence




real-analysis sequences-and-series convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









Martin Sleziak

44.8k10119273




44.8k10119273










asked Mar 7 '16 at 5:56









Mathslover shahMathslover shah

110110




110110












  • $begingroup$
    Making a guess as to what you mean, Baby Rudin 3.28-3.29 would seem helpful.
    $endgroup$
    – ForgotALot
    Mar 7 '16 at 5:59






  • 1




    $begingroup$
    To show divergence if $ale 1$, it is enough (why?) to show divergence when $n=1$. For this use the Integral Test.
    $endgroup$
    – André Nicolas
    Mar 7 '16 at 6:11












  • $begingroup$
    Hello can you make it clear that my question is about n power logn
    $endgroup$
    – Mathslover shah
    Mar 7 '16 at 9:20






  • 3




    $begingroup$
    For some $N$, $n>Nimplieslog(n)>1$. For some other $N$, $n>Nimplieslog(log(n))>1$.
    $endgroup$
    – Yves Daoust
    Mar 7 '16 at 9:24












  • $begingroup$
    Can you edit the problem with n power to the logn please i am not able to do so
    $endgroup$
    – Mathslover shah
    Mar 7 '16 at 9:25


















  • $begingroup$
    Making a guess as to what you mean, Baby Rudin 3.28-3.29 would seem helpful.
    $endgroup$
    – ForgotALot
    Mar 7 '16 at 5:59






  • 1




    $begingroup$
    To show divergence if $ale 1$, it is enough (why?) to show divergence when $n=1$. For this use the Integral Test.
    $endgroup$
    – André Nicolas
    Mar 7 '16 at 6:11












  • $begingroup$
    Hello can you make it clear that my question is about n power logn
    $endgroup$
    – Mathslover shah
    Mar 7 '16 at 9:20






  • 3




    $begingroup$
    For some $N$, $n>Nimplieslog(n)>1$. For some other $N$, $n>Nimplieslog(log(n))>1$.
    $endgroup$
    – Yves Daoust
    Mar 7 '16 at 9:24












  • $begingroup$
    Can you edit the problem with n power to the logn please i am not able to do so
    $endgroup$
    – Mathslover shah
    Mar 7 '16 at 9:25
















$begingroup$
Making a guess as to what you mean, Baby Rudin 3.28-3.29 would seem helpful.
$endgroup$
– ForgotALot
Mar 7 '16 at 5:59




$begingroup$
Making a guess as to what you mean, Baby Rudin 3.28-3.29 would seem helpful.
$endgroup$
– ForgotALot
Mar 7 '16 at 5:59




1




1




$begingroup$
To show divergence if $ale 1$, it is enough (why?) to show divergence when $n=1$. For this use the Integral Test.
$endgroup$
– André Nicolas
Mar 7 '16 at 6:11






$begingroup$
To show divergence if $ale 1$, it is enough (why?) to show divergence when $n=1$. For this use the Integral Test.
$endgroup$
– André Nicolas
Mar 7 '16 at 6:11














$begingroup$
Hello can you make it clear that my question is about n power logn
$endgroup$
– Mathslover shah
Mar 7 '16 at 9:20




$begingroup$
Hello can you make it clear that my question is about n power logn
$endgroup$
– Mathslover shah
Mar 7 '16 at 9:20




3




3




$begingroup$
For some $N$, $n>Nimplieslog(n)>1$. For some other $N$, $n>Nimplieslog(log(n))>1$.
$endgroup$
– Yves Daoust
Mar 7 '16 at 9:24






$begingroup$
For some $N$, $n>Nimplieslog(n)>1$. For some other $N$, $n>Nimplieslog(log(n))>1$.
$endgroup$
– Yves Daoust
Mar 7 '16 at 9:24














$begingroup$
Can you edit the problem with n power to the logn please i am not able to do so
$endgroup$
– Mathslover shah
Mar 7 '16 at 9:25




$begingroup$
Can you edit the problem with n power to the logn please i am not able to do so
$endgroup$
– Mathslover shah
Mar 7 '16 at 9:25










2 Answers
2






active

oldest

votes


















1












$begingroup$

For which $a$ does
$sum_{n=2}^infty {1over n^a}
$
converge?
For which $a$ does it diverge?



If this converges,
then
$sum_{n=2}^infty {1over n^alog n}$
will certainly converge
(do you see why?).






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Since the series is a positive one and the general term's sequence is monotonically decreasing to zero, you can use Cauchy's Condensation Test, and:



    $$frac{2^n}{(2^n)^{log2^n}}=frac{2^n}{2^{n^2log2}}=frac1{2^{nleft(nlog2-1right)}}lefrac1{2^n}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Aaaa that is there . Thanks .
      $endgroup$
      – Mathslover shah
      Mar 7 '16 at 10:16










    • $begingroup$
      Thank you to anyone who downvoted/upvoted this. It would really help me if anyone who downvoted could be please kind enough to write down the reason: is something wrong, or unclear? I try to learn of my mistakes...if I recognize them.
      $endgroup$
      – DonAntonio
      Mar 7 '16 at 10:32











    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    For which $a$ does
    $sum_{n=2}^infty {1over n^a}
    $
    converge?
    For which $a$ does it diverge?



    If this converges,
    then
    $sum_{n=2}^infty {1over n^alog n}$
    will certainly converge
    (do you see why?).






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      For which $a$ does
      $sum_{n=2}^infty {1over n^a}
      $
      converge?
      For which $a$ does it diverge?



      If this converges,
      then
      $sum_{n=2}^infty {1over n^alog n}$
      will certainly converge
      (do you see why?).






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        For which $a$ does
        $sum_{n=2}^infty {1over n^a}
        $
        converge?
        For which $a$ does it diverge?



        If this converges,
        then
        $sum_{n=2}^infty {1over n^alog n}$
        will certainly converge
        (do you see why?).






        share|cite|improve this answer









        $endgroup$



        For which $a$ does
        $sum_{n=2}^infty {1over n^a}
        $
        converge?
        For which $a$ does it diverge?



        If this converges,
        then
        $sum_{n=2}^infty {1over n^alog n}$
        will certainly converge
        (do you see why?).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 7 '16 at 6:10









        marty cohenmarty cohen

        74.2k549128




        74.2k549128























            1












            $begingroup$

            Since the series is a positive one and the general term's sequence is monotonically decreasing to zero, you can use Cauchy's Condensation Test, and:



            $$frac{2^n}{(2^n)^{log2^n}}=frac{2^n}{2^{n^2log2}}=frac1{2^{nleft(nlog2-1right)}}lefrac1{2^n}$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Aaaa that is there . Thanks .
              $endgroup$
              – Mathslover shah
              Mar 7 '16 at 10:16










            • $begingroup$
              Thank you to anyone who downvoted/upvoted this. It would really help me if anyone who downvoted could be please kind enough to write down the reason: is something wrong, or unclear? I try to learn of my mistakes...if I recognize them.
              $endgroup$
              – DonAntonio
              Mar 7 '16 at 10:32
















            1












            $begingroup$

            Since the series is a positive one and the general term's sequence is monotonically decreasing to zero, you can use Cauchy's Condensation Test, and:



            $$frac{2^n}{(2^n)^{log2^n}}=frac{2^n}{2^{n^2log2}}=frac1{2^{nleft(nlog2-1right)}}lefrac1{2^n}$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Aaaa that is there . Thanks .
              $endgroup$
              – Mathslover shah
              Mar 7 '16 at 10:16










            • $begingroup$
              Thank you to anyone who downvoted/upvoted this. It would really help me if anyone who downvoted could be please kind enough to write down the reason: is something wrong, or unclear? I try to learn of my mistakes...if I recognize them.
              $endgroup$
              – DonAntonio
              Mar 7 '16 at 10:32














            1












            1








            1





            $begingroup$

            Since the series is a positive one and the general term's sequence is monotonically decreasing to zero, you can use Cauchy's Condensation Test, and:



            $$frac{2^n}{(2^n)^{log2^n}}=frac{2^n}{2^{n^2log2}}=frac1{2^{nleft(nlog2-1right)}}lefrac1{2^n}$$






            share|cite|improve this answer









            $endgroup$



            Since the series is a positive one and the general term's sequence is monotonically decreasing to zero, you can use Cauchy's Condensation Test, and:



            $$frac{2^n}{(2^n)^{log2^n}}=frac{2^n}{2^{n^2log2}}=frac1{2^{nleft(nlog2-1right)}}lefrac1{2^n}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 7 '16 at 9:58









            DonAntonioDonAntonio

            179k1494233




            179k1494233












            • $begingroup$
              Aaaa that is there . Thanks .
              $endgroup$
              – Mathslover shah
              Mar 7 '16 at 10:16










            • $begingroup$
              Thank you to anyone who downvoted/upvoted this. It would really help me if anyone who downvoted could be please kind enough to write down the reason: is something wrong, or unclear? I try to learn of my mistakes...if I recognize them.
              $endgroup$
              – DonAntonio
              Mar 7 '16 at 10:32


















            • $begingroup$
              Aaaa that is there . Thanks .
              $endgroup$
              – Mathslover shah
              Mar 7 '16 at 10:16










            • $begingroup$
              Thank you to anyone who downvoted/upvoted this. It would really help me if anyone who downvoted could be please kind enough to write down the reason: is something wrong, or unclear? I try to learn of my mistakes...if I recognize them.
              $endgroup$
              – DonAntonio
              Mar 7 '16 at 10:32
















            $begingroup$
            Aaaa that is there . Thanks .
            $endgroup$
            – Mathslover shah
            Mar 7 '16 at 10:16




            $begingroup$
            Aaaa that is there . Thanks .
            $endgroup$
            – Mathslover shah
            Mar 7 '16 at 10:16












            $begingroup$
            Thank you to anyone who downvoted/upvoted this. It would really help me if anyone who downvoted could be please kind enough to write down the reason: is something wrong, or unclear? I try to learn of my mistakes...if I recognize them.
            $endgroup$
            – DonAntonio
            Mar 7 '16 at 10:32




            $begingroup$
            Thank you to anyone who downvoted/upvoted this. It would really help me if anyone who downvoted could be please kind enough to write down the reason: is something wrong, or unclear? I try to learn of my mistakes...if I recognize them.
            $endgroup$
            – DonAntonio
            Mar 7 '16 at 10:32


















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