Does $sum_{n=2}^infty frac{1}{nlog(n)}$ converge or diverge? [duplicate]












1















This question already has an answer here:




  • Infinite series $sum _{n=2}^{infty } frac{1}{n log (n)}$

    3 answers



  • Convergence of $sum_{n=3}^infty frac {1}{n ln n}$ [duplicate]

    3 answers




So I know that $sum_{ninmathbb{N}}1/n$ diverges and $sum_{ninmathbb{N}}1/n^2$ converges. What about the series $sum_{n=2}^infty1/n(log(n))$? I'm pretty confident that it diverges but is there a quick justification?






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marked as duplicate by Martin R, Davide Giraudo, amWhy calculus
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 Nov 20 '18 at 23:05


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • Another one: math.stackexchange.com/q/2223275/42969.
    – Martin R
    Nov 20 '18 at 20:53
















1















This question already has an answer here:




  • Infinite series $sum _{n=2}^{infty } frac{1}{n log (n)}$

    3 answers



  • Convergence of $sum_{n=3}^infty frac {1}{n ln n}$ [duplicate]

    3 answers




So I know that $sum_{ninmathbb{N}}1/n$ diverges and $sum_{ninmathbb{N}}1/n^2$ converges. What about the series $sum_{n=2}^infty1/n(log(n))$? I'm pretty confident that it diverges but is there a quick justification?






calculus sequences-and-series







share|cite|improve this question













marked as duplicate by Martin R, Davide Giraudo, amWhy calculus
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 Nov 20 '18 at 23:05


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • Another one: math.stackexchange.com/q/2223275/42969.
    – Martin R
    Nov 20 '18 at 20:53














1












1








1








This question already has an answer here:




  • Infinite series $sum _{n=2}^{infty } frac{1}{n log (n)}$

    3 answers



  • Convergence of $sum_{n=3}^infty frac {1}{n ln n}$ [duplicate]

    3 answers




So I know that $sum_{ninmathbb{N}}1/n$ diverges and $sum_{ninmathbb{N}}1/n^2$ converges. What about the series $sum_{n=2}^infty1/n(log(n))$? I'm pretty confident that it diverges but is there a quick justification?






calculus sequences-and-series







share|cite|improve this question














This question already has an answer here:




  • Infinite series $sum _{n=2}^{infty } frac{1}{n log (n)}$

    3 answers



  • Convergence of $sum_{n=3}^infty frac {1}{n ln n}$ [duplicate]

    3 answers




So I know that $sum_{ninmathbb{N}}1/n$ diverges and $sum_{ninmathbb{N}}1/n^2$ converges. What about the series $sum_{n=2}^infty1/n(log(n))$? I'm pretty confident that it diverges but is there a quick justification?





This question already has an answer here:




  • Infinite series $sum _{n=2}^{infty } frac{1}{n log (n)}$

    3 answers



  • Convergence of $sum_{n=3}^infty frac {1}{n ln n}$ [duplicate]

    3 answers







calculus sequences-and-series




calculus sequences-and-series






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asked Nov 20 '18 at 20:21









Dragonite

1,045420




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 Nov 20 '18 at 23:05


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • Another one: math.stackexchange.com/q/2223275/42969.
    – Martin R
    Nov 20 '18 at 20:53


















  • Another one: math.stackexchange.com/q/2223275/42969.
    – Martin R
    Nov 20 '18 at 20:53
















Another one: math.stackexchange.com/q/2223275/42969.
– Martin R
Nov 20 '18 at 20:53




Another one: math.stackexchange.com/q/2223275/42969.
– Martin R
Nov 20 '18 at 20:53










3 Answers
3






active

oldest

votes


















6














Diverges by Integral test, since $$displaystyle int_2^infty dfrac{dt}{t ln(t)} = int_2^infty dfrac{d}{dt} ln(ln(t))$$ diverges.






share|cite|improve this answer





























    5














    HINT



    By Cauchy condensation test we can consider the condensed series



    $$sum_{n=2}^infty frac{1}{nlog(n)} rightarrowsum_{n=2}^infty frac{2^n}{2^nlog(2^n)}$$



    and the first converges if and only if the second one converges.



    By the same test we can also determine the convergence for the more general case



    $$sum_{n=2}^infty frac{1}{n^alog^b(n)}$$



    known as Bertrand Series.






    share|cite|improve this answer



















    • 1




      Thanks, this is exactly what I was looking for. :-)
      – Dragonite
      Nov 20 '18 at 20:23



















    2














    I'll see if I can
    come up with a discrete version
    of the integral test.



    $begin{array}\
    ln(ln(t+1))-ln(ln(t))
    &=lnleft(dfrac{ln(t+1)}{ln(t))}right)\
    &=lnleft(dfrac{ln(t)+(ln(t+1)-ln(t))}{ln(t))}right)\
    &=lnleft(1+dfrac{ln(t+1)-ln(t)}{ln(t))}right)\
    &=lnleft(1+dfrac{ln(frac{t+1}{t})}{ln(t))}right)\
    &=lnleft(1+dfrac{ln(1+frac{1}{t})}{ln(t))}right)\
    &<lnleft(1+dfrac{frac{1}{t}}{ln(t))}right)
    qquadtext{since } ln(1+x) < x text{ for } 0 < x\
    &=lnleft(1+dfrac{1}{tln(t))}right)\
    &<dfrac{1}{tln(t))}\
    end{array}
    $



    so



    $begin{array}\
    sum_{k=2}^n dfrac{1}{kln(k))}
    &>sum_{k=2}^n (ln(ln(k+1))-ln(ln(k)))\
    &=ln(ln(n+1))-ln(ln(2)))\
    &to infty\
    end{array}
    $



    Yep - that works.



    By iterating this,
    we can show that
    $sum_n dfrac1{nln(n)ln(ln(n))ln(ln(ln(n)))...}
    $
    diverges.



    All that is needed is
    $ln(1+x) < x$.






    share|cite|improve this answer

















    • 1




      Similar approach here: math.stackexchange.com/a/2223354/42969.
      – Martin R
      Nov 20 '18 at 20:58










    • Not at all surprised.
      – marty cohen
      Nov 21 '18 at 4:39


















    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6














    Diverges by Integral test, since $$displaystyle int_2^infty dfrac{dt}{t ln(t)} = int_2^infty dfrac{d}{dt} ln(ln(t))$$ diverges.






    share|cite|improve this answer


























      6














      Diverges by Integral test, since $$displaystyle int_2^infty dfrac{dt}{t ln(t)} = int_2^infty dfrac{d}{dt} ln(ln(t))$$ diverges.






      share|cite|improve this answer
























        6












        6








        6






        Diverges by Integral test, since $$displaystyle int_2^infty dfrac{dt}{t ln(t)} = int_2^infty dfrac{d}{dt} ln(ln(t))$$ diverges.






        share|cite|improve this answer












        Diverges by Integral test, since $$displaystyle int_2^infty dfrac{dt}{t ln(t)} = int_2^infty dfrac{d}{dt} ln(ln(t))$$ diverges.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 20 '18 at 20:23









        Robert Israel

        318k23208457




        318k23208457























            5














            HINT



            By Cauchy condensation test we can consider the condensed series



            $$sum_{n=2}^infty frac{1}{nlog(n)} rightarrowsum_{n=2}^infty frac{2^n}{2^nlog(2^n)}$$



            and the first converges if and only if the second one converges.



            By the same test we can also determine the convergence for the more general case



            $$sum_{n=2}^infty frac{1}{n^alog^b(n)}$$



            known as Bertrand Series.






            share|cite|improve this answer



















            • 1




              Thanks, this is exactly what I was looking for. :-)
              – Dragonite
              Nov 20 '18 at 20:23
















            5














            HINT



            By Cauchy condensation test we can consider the condensed series



            $$sum_{n=2}^infty frac{1}{nlog(n)} rightarrowsum_{n=2}^infty frac{2^n}{2^nlog(2^n)}$$



            and the first converges if and only if the second one converges.



            By the same test we can also determine the convergence for the more general case



            $$sum_{n=2}^infty frac{1}{n^alog^b(n)}$$



            known as Bertrand Series.






            share|cite|improve this answer



















            • 1




              Thanks, this is exactly what I was looking for. :-)
              – Dragonite
              Nov 20 '18 at 20:23














            5












            5








            5






            HINT



            By Cauchy condensation test we can consider the condensed series



            $$sum_{n=2}^infty frac{1}{nlog(n)} rightarrowsum_{n=2}^infty frac{2^n}{2^nlog(2^n)}$$



            and the first converges if and only if the second one converges.



            By the same test we can also determine the convergence for the more general case



            $$sum_{n=2}^infty frac{1}{n^alog^b(n)}$$



            known as Bertrand Series.






            share|cite|improve this answer














            HINT



            By Cauchy condensation test we can consider the condensed series



            $$sum_{n=2}^infty frac{1}{nlog(n)} rightarrowsum_{n=2}^infty frac{2^n}{2^nlog(2^n)}$$



            and the first converges if and only if the second one converges.



            By the same test we can also determine the convergence for the more general case



            $$sum_{n=2}^infty frac{1}{n^alog^b(n)}$$



            known as Bertrand Series.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 20 '18 at 20:24

























            answered Nov 20 '18 at 20:22









            gimusi

            1




            1








            • 1




              Thanks, this is exactly what I was looking for. :-)
              – Dragonite
              Nov 20 '18 at 20:23














            • 1




              Thanks, this is exactly what I was looking for. :-)
              – Dragonite
              Nov 20 '18 at 20:23








            1




            1




            Thanks, this is exactly what I was looking for. :-)
            – Dragonite
            Nov 20 '18 at 20:23




            Thanks, this is exactly what I was looking for. :-)
            – Dragonite
            Nov 20 '18 at 20:23











            2














            I'll see if I can
            come up with a discrete version
            of the integral test.



            $begin{array}\
            ln(ln(t+1))-ln(ln(t))
            &=lnleft(dfrac{ln(t+1)}{ln(t))}right)\
            &=lnleft(dfrac{ln(t)+(ln(t+1)-ln(t))}{ln(t))}right)\
            &=lnleft(1+dfrac{ln(t+1)-ln(t)}{ln(t))}right)\
            &=lnleft(1+dfrac{ln(frac{t+1}{t})}{ln(t))}right)\
            &=lnleft(1+dfrac{ln(1+frac{1}{t})}{ln(t))}right)\
            &<lnleft(1+dfrac{frac{1}{t}}{ln(t))}right)
            qquadtext{since } ln(1+x) < x text{ for } 0 < x\
            &=lnleft(1+dfrac{1}{tln(t))}right)\
            &<dfrac{1}{tln(t))}\
            end{array}
            $



            so



            $begin{array}\
            sum_{k=2}^n dfrac{1}{kln(k))}
            &>sum_{k=2}^n (ln(ln(k+1))-ln(ln(k)))\
            &=ln(ln(n+1))-ln(ln(2)))\
            &to infty\
            end{array}
            $



            Yep - that works.



            By iterating this,
            we can show that
            $sum_n dfrac1{nln(n)ln(ln(n))ln(ln(ln(n)))...}
            $
            diverges.



            All that is needed is
            $ln(1+x) < x$.






            share|cite|improve this answer

















            • 1




              Similar approach here: math.stackexchange.com/a/2223354/42969.
              – Martin R
              Nov 20 '18 at 20:58










            • Not at all surprised.
              – marty cohen
              Nov 21 '18 at 4:39
















            2














            I'll see if I can
            come up with a discrete version
            of the integral test.



            $begin{array}\
            ln(ln(t+1))-ln(ln(t))
            &=lnleft(dfrac{ln(t+1)}{ln(t))}right)\
            &=lnleft(dfrac{ln(t)+(ln(t+1)-ln(t))}{ln(t))}right)\
            &=lnleft(1+dfrac{ln(t+1)-ln(t)}{ln(t))}right)\
            &=lnleft(1+dfrac{ln(frac{t+1}{t})}{ln(t))}right)\
            &=lnleft(1+dfrac{ln(1+frac{1}{t})}{ln(t))}right)\
            &<lnleft(1+dfrac{frac{1}{t}}{ln(t))}right)
            qquadtext{since } ln(1+x) < x text{ for } 0 < x\
            &=lnleft(1+dfrac{1}{tln(t))}right)\
            &<dfrac{1}{tln(t))}\
            end{array}
            $



            so



            $begin{array}\
            sum_{k=2}^n dfrac{1}{kln(k))}
            &>sum_{k=2}^n (ln(ln(k+1))-ln(ln(k)))\
            &=ln(ln(n+1))-ln(ln(2)))\
            &to infty\
            end{array}
            $



            Yep - that works.



            By iterating this,
            we can show that
            $sum_n dfrac1{nln(n)ln(ln(n))ln(ln(ln(n)))...}
            $
            diverges.



            All that is needed is
            $ln(1+x) < x$.






            share|cite|improve this answer

















            • 1




              Similar approach here: math.stackexchange.com/a/2223354/42969.
              – Martin R
              Nov 20 '18 at 20:58










            • Not at all surprised.
              – marty cohen
              Nov 21 '18 at 4:39














            2












            2








            2






            I'll see if I can
            come up with a discrete version
            of the integral test.



            $begin{array}\
            ln(ln(t+1))-ln(ln(t))
            &=lnleft(dfrac{ln(t+1)}{ln(t))}right)\
            &=lnleft(dfrac{ln(t)+(ln(t+1)-ln(t))}{ln(t))}right)\
            &=lnleft(1+dfrac{ln(t+1)-ln(t)}{ln(t))}right)\
            &=lnleft(1+dfrac{ln(frac{t+1}{t})}{ln(t))}right)\
            &=lnleft(1+dfrac{ln(1+frac{1}{t})}{ln(t))}right)\
            &<lnleft(1+dfrac{frac{1}{t}}{ln(t))}right)
            qquadtext{since } ln(1+x) < x text{ for } 0 < x\
            &=lnleft(1+dfrac{1}{tln(t))}right)\
            &<dfrac{1}{tln(t))}\
            end{array}
            $



            so



            $begin{array}\
            sum_{k=2}^n dfrac{1}{kln(k))}
            &>sum_{k=2}^n (ln(ln(k+1))-ln(ln(k)))\
            &=ln(ln(n+1))-ln(ln(2)))\
            &to infty\
            end{array}
            $



            Yep - that works.



            By iterating this,
            we can show that
            $sum_n dfrac1{nln(n)ln(ln(n))ln(ln(ln(n)))...}
            $
            diverges.



            All that is needed is
            $ln(1+x) < x$.






            share|cite|improve this answer












            I'll see if I can
            come up with a discrete version
            of the integral test.



            $begin{array}\
            ln(ln(t+1))-ln(ln(t))
            &=lnleft(dfrac{ln(t+1)}{ln(t))}right)\
            &=lnleft(dfrac{ln(t)+(ln(t+1)-ln(t))}{ln(t))}right)\
            &=lnleft(1+dfrac{ln(t+1)-ln(t)}{ln(t))}right)\
            &=lnleft(1+dfrac{ln(frac{t+1}{t})}{ln(t))}right)\
            &=lnleft(1+dfrac{ln(1+frac{1}{t})}{ln(t))}right)\
            &<lnleft(1+dfrac{frac{1}{t}}{ln(t))}right)
            qquadtext{since } ln(1+x) < x text{ for } 0 < x\
            &=lnleft(1+dfrac{1}{tln(t))}right)\
            &<dfrac{1}{tln(t))}\
            end{array}
            $



            so



            $begin{array}\
            sum_{k=2}^n dfrac{1}{kln(k))}
            &>sum_{k=2}^n (ln(ln(k+1))-ln(ln(k)))\
            &=ln(ln(n+1))-ln(ln(2)))\
            &to infty\
            end{array}
            $



            Yep - that works.



            By iterating this,
            we can show that
            $sum_n dfrac1{nln(n)ln(ln(n))ln(ln(ln(n)))...}
            $
            diverges.



            All that is needed is
            $ln(1+x) < x$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 20 '18 at 20:48









            marty cohen

            72.6k549127




            72.6k549127








            • 1




              Similar approach here: math.stackexchange.com/a/2223354/42969.
              – Martin R
              Nov 20 '18 at 20:58










            • Not at all surprised.
              – marty cohen
              Nov 21 '18 at 4:39














            • 1




              Similar approach here: math.stackexchange.com/a/2223354/42969.
              – Martin R
              Nov 20 '18 at 20:58










            • Not at all surprised.
              – marty cohen
              Nov 21 '18 at 4:39








            1




            1




            Similar approach here: math.stackexchange.com/a/2223354/42969.
            – Martin R
            Nov 20 '18 at 20:58




            Similar approach here: math.stackexchange.com/a/2223354/42969.
            – Martin R
            Nov 20 '18 at 20:58












            Not at all surprised.
            – marty cohen
            Nov 21 '18 at 4:39




            Not at all surprised.
            – marty cohen
            Nov 21 '18 at 4:39



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