The following series converge or diverge?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
4
down vote

favorite












Does the following series coverge or diverge?



$$sum_n=2^infty frac1(log n)^loglog n$$



My attempts : I know that $log(log n) > 2$



So $displaystylesum_n=2^inftyfrac1(log n)^loglog nlesum_n=2^infty frac1(log n)^2lesumfrac1n^2$



so the given series converges.



Is it correct?




real-analysis




share|cite|improve this question


















  • 1




    Check your inequalities, $ln n<n$ !!!
    – Nicolas FRANCOIS
    Aug 8 at 8:29










  • Since the terms are decreasing you can use the integral test. A simple substitution ($y=log x)$ will make it clear that the integral diverges.
    – Kavi Rama Murthy
    Aug 8 at 8:35














up vote
4
down vote

favorite












Does the following series coverge or diverge?



$$sum_n=2^infty frac1(log n)^loglog n$$



My attempts : I know that $log(log n) > 2$



So $displaystylesum_n=2^inftyfrac1(log n)^loglog nlesum_n=2^infty frac1(log n)^2lesumfrac1n^2$



so the given series converges.



Is it correct?




real-analysis




share|cite|improve this question


















  • 1




    Check your inequalities, $ln n<n$ !!!
    – Nicolas FRANCOIS
    Aug 8 at 8:29










  • Since the terms are decreasing you can use the integral test. A simple substitution ($y=log x)$ will make it clear that the integral diverges.
    – Kavi Rama Murthy
    Aug 8 at 8:35












up vote
4
down vote

favorite









up vote
4
down vote

favorite











Does the following series coverge or diverge?



$$sum_n=2^infty frac1(log n)^loglog n$$



My attempts : I know that $log(log n) > 2$



So $displaystylesum_n=2^inftyfrac1(log n)^loglog nlesum_n=2^infty frac1(log n)^2lesumfrac1n^2$



so the given series converges.



Is it correct?




real-analysis




share|cite|improve this question














Does the following series coverge or diverge?



$$sum_n=2^infty frac1(log n)^loglog n$$



My attempts : I know that $log(log n) > 2$



So $displaystylesum_n=2^inftyfrac1(log n)^loglog nlesum_n=2^infty frac1(log n)^2lesumfrac1n^2$



so the given series converges.



Is it correct?





real-analysis





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 8 at 8:49









Bernard

111k635103




111k635103










asked Aug 8 at 8:27









stupid

646111




646111







  • 1




    Check your inequalities, $ln n<n$ !!!
    – Nicolas FRANCOIS
    Aug 8 at 8:29










  • Since the terms are decreasing you can use the integral test. A simple substitution ($y=log x)$ will make it clear that the integral diverges.
    – Kavi Rama Murthy
    Aug 8 at 8:35












  • 1




    Check your inequalities, $ln n<n$ !!!
    – Nicolas FRANCOIS
    Aug 8 at 8:29










  • Since the terms are decreasing you can use the integral test. A simple substitution ($y=log x)$ will make it clear that the integral diverges.
    – Kavi Rama Murthy
    Aug 8 at 8:35







1




1




Check your inequalities, $ln n<n$ !!!
– Nicolas FRANCOIS
Aug 8 at 8:29




Check your inequalities, $ln n<n$ !!!
– Nicolas FRANCOIS
Aug 8 at 8:29












Since the terms are decreasing you can use the integral test. A simple substitution ($y=log x)$ will make it clear that the integral diverges.
– Kavi Rama Murthy
Aug 8 at 8:35




Since the terms are decreasing you can use the integral test. A simple substitution ($y=log x)$ will make it clear that the integral diverges.
– Kavi Rama Murthy
Aug 8 at 8:35










2 Answers
2






active

oldest

votes

















up vote
7
down vote



accepted










It's incorrect. We don't have $log n ge n$.



Instead, $(log n)^log log n = exp((log log n)^2) le exp(((log n)^1/2)^2) = n$ for $n ge N_0$, so $$sum_n=2^infty frac 1 (log n)^log log n ge C + sum_n=N_0^infty frac 1 n = infty$$



Implicit result used: $log x le x^varepsilon$ for sufficiently large $x$ (depending on $varepsilon$) for any $varepsilon > 0$.






share|cite|improve this answer



























    up vote
    3
    down vote













    The series is divergent. Note that $a_n= frac1(log n)^loglog n=exp(-(loglog n)^2)$ is positive and decreasing and, by Cauchy condensation test, the convergence of the given series is equivalent to the convergence of
    $$sum_n=2^infty 2^nexp(-(loglog 2^n)^2)=sum_n=2^infty exp(nlog 2-(log n+loglog 2)^2)$$
    which is divergent because $nlog 2-(log n+loglog 2)^2to +infty$.






    share|cite|improve this answer






















      Your Answer




      StackExchange.ifUsing("editor", function ()
      return StackExchange.using("mathjaxEditing", function ()
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      );
      );
      , "mathjax-editing");

      StackExchange.ready(function()
      var channelOptions =
      tags: "".split(" "),
      id: "69"
      ;
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function()
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled)
      StackExchange.using("snippets", function()
      createEditor();
      );

      else
      createEditor();

      );

      function createEditor()
      StackExchange.prepareEditor(
      heartbeatType: 'answer',
      convertImagesToLinks: true,
      noModals: false,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      );



      );













       

      draft saved


      draft discarded


















      StackExchange.ready(
      function ()
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2875783%2fthe-following-series-converge-or-diverge%23new-answer', 'question_page');

      );

      Post as a guest

















      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      7
      down vote



      accepted










      It's incorrect. We don't have $log n ge n$.



      Instead, $(log n)^log log n = exp((log log n)^2) le exp(((log n)^1/2)^2) = n$ for $n ge N_0$, so $$sum_n=2^infty frac 1 (log n)^log log n ge C + sum_n=N_0^infty frac 1 n = infty$$



      Implicit result used: $log x le x^varepsilon$ for sufficiently large $x$ (depending on $varepsilon$) for any $varepsilon > 0$.






      share|cite|improve this answer
























        up vote
        7
        down vote



        accepted










        It's incorrect. We don't have $log n ge n$.



        Instead, $(log n)^log log n = exp((log log n)^2) le exp(((log n)^1/2)^2) = n$ for $n ge N_0$, so $$sum_n=2^infty frac 1 (log n)^log log n ge C + sum_n=N_0^infty frac 1 n = infty$$



        Implicit result used: $log x le x^varepsilon$ for sufficiently large $x$ (depending on $varepsilon$) for any $varepsilon > 0$.






        share|cite|improve this answer






















          up vote
          7
          down vote



          accepted







          up vote
          7
          down vote



          accepted






          It's incorrect. We don't have $log n ge n$.



          Instead, $(log n)^log log n = exp((log log n)^2) le exp(((log n)^1/2)^2) = n$ for $n ge N_0$, so $$sum_n=2^infty frac 1 (log n)^log log n ge C + sum_n=N_0^infty frac 1 n = infty$$



          Implicit result used: $log x le x^varepsilon$ for sufficiently large $x$ (depending on $varepsilon$) for any $varepsilon > 0$.






          share|cite|improve this answer












          It's incorrect. We don't have $log n ge n$.



          Instead, $(log n)^log log n = exp((log log n)^2) le exp(((log n)^1/2)^2) = n$ for $n ge N_0$, so $$sum_n=2^infty frac 1 (log n)^log log n ge C + sum_n=N_0^infty frac 1 n = infty$$



          Implicit result used: $log x le x^varepsilon$ for sufficiently large $x$ (depending on $varepsilon$) for any $varepsilon > 0$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 8 at 8:32









          Kenny Lau

          19k2157




          19k2157




















              up vote
              3
              down vote













              The series is divergent. Note that $a_n= frac1(log n)^loglog n=exp(-(loglog n)^2)$ is positive and decreasing and, by Cauchy condensation test, the convergence of the given series is equivalent to the convergence of
              $$sum_n=2^infty 2^nexp(-(loglog 2^n)^2)=sum_n=2^infty exp(nlog 2-(log n+loglog 2)^2)$$
              which is divergent because $nlog 2-(log n+loglog 2)^2to +infty$.






              share|cite|improve this answer


























                up vote
                3
                down vote













                The series is divergent. Note that $a_n= frac1(log n)^loglog n=exp(-(loglog n)^2)$ is positive and decreasing and, by Cauchy condensation test, the convergence of the given series is equivalent to the convergence of
                $$sum_n=2^infty 2^nexp(-(loglog 2^n)^2)=sum_n=2^infty exp(nlog 2-(log n+loglog 2)^2)$$
                which is divergent because $nlog 2-(log n+loglog 2)^2to +infty$.






                share|cite|improve this answer
























                  up vote
                  3
                  down vote










                  up vote
                  3
                  down vote









                  The series is divergent. Note that $a_n= frac1(log n)^loglog n=exp(-(loglog n)^2)$ is positive and decreasing and, by Cauchy condensation test, the convergence of the given series is equivalent to the convergence of
                  $$sum_n=2^infty 2^nexp(-(loglog 2^n)^2)=sum_n=2^infty exp(nlog 2-(log n+loglog 2)^2)$$
                  which is divergent because $nlog 2-(log n+loglog 2)^2to +infty$.






                  share|cite|improve this answer














                  The series is divergent. Note that $a_n= frac1(log n)^loglog n=exp(-(loglog n)^2)$ is positive and decreasing and, by Cauchy condensation test, the convergence of the given series is equivalent to the convergence of
                  $$sum_n=2^infty 2^nexp(-(loglog 2^n)^2)=sum_n=2^infty exp(nlog 2-(log n+loglog 2)^2)$$
                  which is divergent because $nlog 2-(log n+loglog 2)^2to +infty$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Aug 8 at 9:00

























                  answered Aug 8 at 8:55









                  Robert Z

                  85.1k1055123




                  85.1k1055123



























                       

                      draft saved


                      draft discarded















































                       


                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function ()
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2875783%2fthe-following-series-converge-or-diverge%23new-answer', 'question_page');

                      );

                      Post as a guest
































































                      Comments

                      Post a Comment

                      Popular posts from this blog

                      What does second last employer means? [closed]

                      Image
                      Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote -3 down vote favorite Say, for example I have worked in the following companies, Company A => 2010 - 2011 Company B => 2011 - 2012 Company C => 2012 - Till Date Company D => Have offer Now assume Company D is asking to submit documents from my second last employer. So from the list of companies given, which company would be qualified as second last employer? software-industry employer share | improve this question asked Aug 10 '14 at 5:16 Rajaprabhu Aravindasamy 559 1 6 13 closed as off-topic by Jim G., jmort253 ♦ Aug 11 '14 at 1:58 This question appears to be off-topic. The users who voted to close gave this specific reason: "Questions seeking advice on company-specific regulations, agreements, or policies should be directed to your manager or HR department. Questions that address...
                      Read more

                      List of Gilmore Girls characters

                      Image
                      The season three cast—(Back row): Lane, Michel, Paris, Emily, Richard, Sookie, Miss Patty, Kirk; Front row: Jess, Luke, Lorelai, Rory, Dean This is a list of characters for the comedy-drama television series Gilmore Girls . Contents 1 Main characters 1.1 Lorelai Gilmore 1.2 Rory Gilmore 1.3 Sookie St. James 1.4 Lane Kim 1.5 Michel Gerard 1.6 Luke Danes 1.7 Emily Gilmore 1.8 Richard Gilmore 1.9 Paris Geller 1.10 Dean Forester 1.11 Jess Mariano 1.12 Kirk Gleason 1.13 Jason Stiles 1.14 Logan Huntzberger 2 Major recurring characters 2.1 Christopher Hayden 2.2 Jackson Belleville 2.3 Mrs. Kim 2.4 Tristin DuGray 2.5 Max Medina 2.6 Taylor Doose 2.7 Dave Rygalski 2.8 Marty 3 Recurring characters 3.1 Stars Hollow 3.2 Chilton 3.3 Yale 3.4 Other 4 Notable guest stars 5 Unseen characters 6 Notes 7 References Main characters Actor Character Appearances Season 1 Season 2 Season 3 Season 4 Season 5 Season 6 Season 7 A Year in the Life Lauren Graham Lorelai Gilmore...
                      Read more

                      One-line joke

                      A one-liner is a joke that is delivered in a single line. A good one-liner is said to be pithy - concise and meaningful. [1] Comedians and actors use this comedic method as part of their act, e.g. Jimmy Carr, Tommy Cooper, Rodney Dangerfield, Norm Macdonald, Ken Dodd, Stewart Francis, Zach Galifianakis, Mitch Hedberg, Anthony Jeselnik, Milton Jones, Shappi Khorsandi, Jay London, Mark Linn-Baker, Demetri Martin, Groucho Marx, Dan Mintz, Emo Philips, Tim Vine, Steven Wright and Henny Youngman and Arnold Schwarzenegger. [ citation needed ] Many fictional characters are also known to deliver one-liners, including James Bond, who usually includes pithy and laconic quips after disposing of a villain. [ citation needed ] Examples "A baby seal walks into a club." "A dyslexic man walks into a bra." (Anonymous) "There are three types of people, those who can count and those who can't." "Venison's dear, isn't it?" (Jimmy Carr) "An escala...
                      Read more