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




real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 8 at 8:49









Bernard

112k636104




112k636104










asked Aug 8 at 8:27









jasmine

843113




843113







  • 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

          19.1k2158




          19.1k2158




















              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

                  85.9k1056124




                  85.9k1056124



























                       

                      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
































































                      -

                      Popular posts from this blog

                      Which professions warranted travel in Medieval times?

                      Image
                      Clash Royale CLAN TAG #URR8PPP up vote 54 down vote favorite 15 I'm trying to come up with characters whose backgrounds allow them to have homes, histories, and skills, but will still enable or encourage them to travel in a typical medieval setting. I've never really liked "Adventurer" as a job, and would like my story to have a kind of unlikely-hero vibe. besides the common "Traveling merchant", what kinds of artisans or professionals in fantasy would have a reason to move around the world (or at least visit other settlements)? First time posting, so let me know what to improve. Thanks! EDIT: When I say traveling, I was thinking someone who spends multiple days/weeks/months away from their place of origin as they go elsewhere for their business, or to gather materials for it. medieval travel share | improve this question edited Aug 1 at 15:16 asked Aug 1 at 15:07 Julia Witham 279 2 7 11 Please ...
                      Read more

                      How do so many people here on Academia.SE, and in general, afford lavish higher education programs?

                      Image
                      Clash Royale CLAN TAG #URR8PPP up vote 45 down vote favorite 7 I graduated last year with a BS in Computer Science. I do not come from a particularly wealthy family, (very low middle class), and was only able to afford my education via school loans/scholarships and working low-wage jobs, with no support from my family or anyone else. I realize some scholarship/loan programs extend into grad school. But I don't think those cover all costs. Now I'm lucky enough to have a decent-paying salary in industry, along with benefits and living on my own, and I'm thinking one day (sooner rather than later) I'd like to try my hand at research or continue my degree, but these things cost money, in addition to my regular living expenses. I read posts here often about academic people being in positions of choosing between these crazy research grad programs, some in Japan, or Ireland (while they currently live in the U.S./U.K.) then moving around between countries/universit...
                      Read more

                      Trouble downloading packages list due to a “Hash sum mismatch” error

                      Image
                      Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote 331 down vote favorite 117 When I check for updates, I get a "Failed To Download Repository Information" error. This is what comes up under details: W: Failed to fetch gzip:/var/lib/apt/lists/partial/us.archive.ubuntu.com_ubuntu_dists_natty_main_source_Sources Hash Sum mismatch, E: Some index files failed to download. They have been ignored, or old ones used instead. apt share | improve this question edited Nov 12 '14 at 23:37 Braiam 49.5k 19 130 210 asked May 9 '11 at 20:55 Rob 5,300 10 26 40 add a comment  |  up vote 331 down vote favorite 117 When I check for updates, I get a "Failed To Download Repository Information" error. This is what comes up under details: W: Failed to fetch gzip:/var/lib/apt/lists/partial/us.archive.ubuntu.com_ubuntu_dists_nat...
                      Read more