Convergence of the series $ sum_{n=1}^{+ infty } frac{1}{n cdot ln(n) cdot (ln(ln(n)))^{a}} $












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I have a problem with a convergence of the series $ sum_{n=1}^{+ infty } frac{1}{n cdot ln(n) cdot (ln(ln(n)))^{a}}$ depending on the parameter $a$. I know that $frac{1}{n} rightarrow 0$, $frac{1}{ln(n)} rightarrow 0$, but I don't know how to deal with $frac{1}{ln(ln(n)))^{a}}$ and how to get to the final answer. So far I solved tasks with $frac{1}{n ^{a}}$ and I know that $a>1 Leftrightarrow sum_{n=1}^{+ infty }frac{1}{n ^{a}}$ is convergent, but in this task I cannot use it. Can you help me?






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    2












    $begingroup$


    I have a problem with a convergence of the series $ sum_{n=1}^{+ infty } frac{1}{n cdot ln(n) cdot (ln(ln(n)))^{a}}$ depending on the parameter $a$. I know that $frac{1}{n} rightarrow 0$, $frac{1}{ln(n)} rightarrow 0$, but I don't know how to deal with $frac{1}{ln(ln(n)))^{a}}$ and how to get to the final answer. So far I solved tasks with $frac{1}{n ^{a}}$ and I know that $a>1 Leftrightarrow sum_{n=1}^{+ infty }frac{1}{n ^{a}}$ is convergent, but in this task I cannot use it. Can you help me?






    real-analysis







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I have a problem with a convergence of the series $ sum_{n=1}^{+ infty } frac{1}{n cdot ln(n) cdot (ln(ln(n)))^{a}}$ depending on the parameter $a$. I know that $frac{1}{n} rightarrow 0$, $frac{1}{ln(n)} rightarrow 0$, but I don't know how to deal with $frac{1}{ln(ln(n)))^{a}}$ and how to get to the final answer. So far I solved tasks with $frac{1}{n ^{a}}$ and I know that $a>1 Leftrightarrow sum_{n=1}^{+ infty }frac{1}{n ^{a}}$ is convergent, but in this task I cannot use it. Can you help me?






      real-analysis







      share|cite|improve this question











      $endgroup$




      I have a problem with a convergence of the series $ sum_{n=1}^{+ infty } frac{1}{n cdot ln(n) cdot (ln(ln(n)))^{a}}$ depending on the parameter $a$. I know that $frac{1}{n} rightarrow 0$, $frac{1}{ln(n)} rightarrow 0$, but I don't know how to deal with $frac{1}{ln(ln(n)))^{a}}$ and how to get to the final answer. So far I solved tasks with $frac{1}{n ^{a}}$ and I know that $a>1 Leftrightarrow sum_{n=1}^{+ infty }frac{1}{n ^{a}}$ is convergent, but in this task I cannot use it. Can you help me?






      real-analysis




      real-analysis






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 13 '18 at 21:25







      MP3129

















      asked Dec 13 '18 at 19:46









      MP3129MP3129

      36010




      36010






















          2 Answers
          2






          active

          oldest

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          3












          $begingroup$

          This is set up very naturally for the integral test, after noting that the substitution $u = ln(ln x)$ leads to



          $$int frac{1}{x ln x (ln ln x)^{a}} , dx = int frac 1 {u^a} , du.$$



          The natural caveat, however, is that your series shouldn't start at $n = 1$ because $ln1 $ is a problem, and that $1/n to 0$ not $to 1$.






          share|cite|improve this answer









          $endgroup$





















            3












            $begingroup$

            By Cauchy condensation test we can check the convergence by the condensed series $sum 2^na_{2^n}$ that is



            $$sum_{n=1}^{ infty } frac{2^n}{2^n cdot ln(2^n) cdot (ln(ln(2^n)))^{a}}=sum_{n=1}^{ infty } frac{1}{ nln(2) cdot (ln (n ln 2))^{a}}$$



            and then we can apply the test again to obtain



            $$sum_{n=1}^{ infty }frac{1}{ln(2)cdot(ln(2^{n}ln(2))^{a}} =sum_{n=1}^{ infty }frac{1}{ln(2)cdot(ln(2^{n})+ln(ln(2)) )^{a}} =$$$$=sum_{n=1}^{ infty }frac{1}{ln(2)cdot n^a(ln(2)+ln(ln(2))/n )^{a}}$$



            and for the latter we can use limit comparison test with $sum_{n=1}^{ infty }frac1{n^a}$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Ok, so I have a $frac{1}{ln(2)cdot(ln(2^{n}ln(2))^{a}}$ and $ln(2) rightarrow ln(2) $, więc we have to deal only $(ln(2^{n}ln(2))^{a}$, but I still do not know how to do it
              $endgroup$
              – MP3129
              Dec 13 '18 at 20:10










            • $begingroup$
              @MP3129 You are almost done indeed $$frac{1}{ln(2)cdot(ln(2^{n}ln(2))^{a}} =frac{1}{ln(2)cdot(ln(2^{n})+ln(ln(2)) )^{a}} =frac{1}{ln(2)cdot n^a(ln(2)+ln(ln(2))/n )^{a}}$$
              $endgroup$
              – gimusi
              Dec 13 '18 at 20:27












            • $begingroup$
              But I don't know what I can do with $(ln(2)+ln(ln(2))/n)^{a}$
              $endgroup$
              – MP3129
              Dec 13 '18 at 21:05










            • $begingroup$
              @MP3129 Simply note that $(ln(2)+ln(ln(2))/n)^{a}to (ln 2)^a$ then for the series we can refer to limit comparison test with $sum frac1{n^a}$.
              $endgroup$
              – gimusi
              Dec 13 '18 at 21:07











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            2 Answers
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            2 Answers
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            active

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            active

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            3












            $begingroup$

            This is set up very naturally for the integral test, after noting that the substitution $u = ln(ln x)$ leads to



            $$int frac{1}{x ln x (ln ln x)^{a}} , dx = int frac 1 {u^a} , du.$$



            The natural caveat, however, is that your series shouldn't start at $n = 1$ because $ln1 $ is a problem, and that $1/n to 0$ not $to 1$.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              This is set up very naturally for the integral test, after noting that the substitution $u = ln(ln x)$ leads to



              $$int frac{1}{x ln x (ln ln x)^{a}} , dx = int frac 1 {u^a} , du.$$



              The natural caveat, however, is that your series shouldn't start at $n = 1$ because $ln1 $ is a problem, and that $1/n to 0$ not $to 1$.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                This is set up very naturally for the integral test, after noting that the substitution $u = ln(ln x)$ leads to



                $$int frac{1}{x ln x (ln ln x)^{a}} , dx = int frac 1 {u^a} , du.$$



                The natural caveat, however, is that your series shouldn't start at $n = 1$ because $ln1 $ is a problem, and that $1/n to 0$ not $to 1$.






                share|cite|improve this answer









                $endgroup$



                This is set up very naturally for the integral test, after noting that the substitution $u = ln(ln x)$ leads to



                $$int frac{1}{x ln x (ln ln x)^{a}} , dx = int frac 1 {u^a} , du.$$



                The natural caveat, however, is that your series shouldn't start at $n = 1$ because $ln1 $ is a problem, and that $1/n to 0$ not $to 1$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 13 '18 at 19:49









                T. BongersT. Bongers

                23.5k54762




                23.5k54762























                    3












                    $begingroup$

                    By Cauchy condensation test we can check the convergence by the condensed series $sum 2^na_{2^n}$ that is



                    $$sum_{n=1}^{ infty } frac{2^n}{2^n cdot ln(2^n) cdot (ln(ln(2^n)))^{a}}=sum_{n=1}^{ infty } frac{1}{ nln(2) cdot (ln (n ln 2))^{a}}$$



                    and then we can apply the test again to obtain



                    $$sum_{n=1}^{ infty }frac{1}{ln(2)cdot(ln(2^{n}ln(2))^{a}} =sum_{n=1}^{ infty }frac{1}{ln(2)cdot(ln(2^{n})+ln(ln(2)) )^{a}} =$$$$=sum_{n=1}^{ infty }frac{1}{ln(2)cdot n^a(ln(2)+ln(ln(2))/n )^{a}}$$



                    and for the latter we can use limit comparison test with $sum_{n=1}^{ infty }frac1{n^a}$.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Ok, so I have a $frac{1}{ln(2)cdot(ln(2^{n}ln(2))^{a}}$ and $ln(2) rightarrow ln(2) $, więc we have to deal only $(ln(2^{n}ln(2))^{a}$, but I still do not know how to do it
                      $endgroup$
                      – MP3129
                      Dec 13 '18 at 20:10










                    • $begingroup$
                      @MP3129 You are almost done indeed $$frac{1}{ln(2)cdot(ln(2^{n}ln(2))^{a}} =frac{1}{ln(2)cdot(ln(2^{n})+ln(ln(2)) )^{a}} =frac{1}{ln(2)cdot n^a(ln(2)+ln(ln(2))/n )^{a}}$$
                      $endgroup$
                      – gimusi
                      Dec 13 '18 at 20:27












                    • $begingroup$
                      But I don't know what I can do with $(ln(2)+ln(ln(2))/n)^{a}$
                      $endgroup$
                      – MP3129
                      Dec 13 '18 at 21:05










                    • $begingroup$
                      @MP3129 Simply note that $(ln(2)+ln(ln(2))/n)^{a}to (ln 2)^a$ then for the series we can refer to limit comparison test with $sum frac1{n^a}$.
                      $endgroup$
                      – gimusi
                      Dec 13 '18 at 21:07
















                    3












                    $begingroup$

                    By Cauchy condensation test we can check the convergence by the condensed series $sum 2^na_{2^n}$ that is



                    $$sum_{n=1}^{ infty } frac{2^n}{2^n cdot ln(2^n) cdot (ln(ln(2^n)))^{a}}=sum_{n=1}^{ infty } frac{1}{ nln(2) cdot (ln (n ln 2))^{a}}$$



                    and then we can apply the test again to obtain



                    $$sum_{n=1}^{ infty }frac{1}{ln(2)cdot(ln(2^{n}ln(2))^{a}} =sum_{n=1}^{ infty }frac{1}{ln(2)cdot(ln(2^{n})+ln(ln(2)) )^{a}} =$$$$=sum_{n=1}^{ infty }frac{1}{ln(2)cdot n^a(ln(2)+ln(ln(2))/n )^{a}}$$



                    and for the latter we can use limit comparison test with $sum_{n=1}^{ infty }frac1{n^a}$.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Ok, so I have a $frac{1}{ln(2)cdot(ln(2^{n}ln(2))^{a}}$ and $ln(2) rightarrow ln(2) $, więc we have to deal only $(ln(2^{n}ln(2))^{a}$, but I still do not know how to do it
                      $endgroup$
                      – MP3129
                      Dec 13 '18 at 20:10










                    • $begingroup$
                      @MP3129 You are almost done indeed $$frac{1}{ln(2)cdot(ln(2^{n}ln(2))^{a}} =frac{1}{ln(2)cdot(ln(2^{n})+ln(ln(2)) )^{a}} =frac{1}{ln(2)cdot n^a(ln(2)+ln(ln(2))/n )^{a}}$$
                      $endgroup$
                      – gimusi
                      Dec 13 '18 at 20:27












                    • $begingroup$
                      But I don't know what I can do with $(ln(2)+ln(ln(2))/n)^{a}$
                      $endgroup$
                      – MP3129
                      Dec 13 '18 at 21:05










                    • $begingroup$
                      @MP3129 Simply note that $(ln(2)+ln(ln(2))/n)^{a}to (ln 2)^a$ then for the series we can refer to limit comparison test with $sum frac1{n^a}$.
                      $endgroup$
                      – gimusi
                      Dec 13 '18 at 21:07














                    3












                    3








                    3





                    $begingroup$

                    By Cauchy condensation test we can check the convergence by the condensed series $sum 2^na_{2^n}$ that is



                    $$sum_{n=1}^{ infty } frac{2^n}{2^n cdot ln(2^n) cdot (ln(ln(2^n)))^{a}}=sum_{n=1}^{ infty } frac{1}{ nln(2) cdot (ln (n ln 2))^{a}}$$



                    and then we can apply the test again to obtain



                    $$sum_{n=1}^{ infty }frac{1}{ln(2)cdot(ln(2^{n}ln(2))^{a}} =sum_{n=1}^{ infty }frac{1}{ln(2)cdot(ln(2^{n})+ln(ln(2)) )^{a}} =$$$$=sum_{n=1}^{ infty }frac{1}{ln(2)cdot n^a(ln(2)+ln(ln(2))/n )^{a}}$$



                    and for the latter we can use limit comparison test with $sum_{n=1}^{ infty }frac1{n^a}$.






                    share|cite|improve this answer











                    $endgroup$



                    By Cauchy condensation test we can check the convergence by the condensed series $sum 2^na_{2^n}$ that is



                    $$sum_{n=1}^{ infty } frac{2^n}{2^n cdot ln(2^n) cdot (ln(ln(2^n)))^{a}}=sum_{n=1}^{ infty } frac{1}{ nln(2) cdot (ln (n ln 2))^{a}}$$



                    and then we can apply the test again to obtain



                    $$sum_{n=1}^{ infty }frac{1}{ln(2)cdot(ln(2^{n}ln(2))^{a}} =sum_{n=1}^{ infty }frac{1}{ln(2)cdot(ln(2^{n})+ln(ln(2)) )^{a}} =$$$$=sum_{n=1}^{ infty }frac{1}{ln(2)cdot n^a(ln(2)+ln(ln(2))/n )^{a}}$$



                    and for the latter we can use limit comparison test with $sum_{n=1}^{ infty }frac1{n^a}$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 14 '18 at 8:29

























                    answered Dec 13 '18 at 19:49









                    gimusigimusi

                    92.9k84494




                    92.9k84494












                    • $begingroup$
                      Ok, so I have a $frac{1}{ln(2)cdot(ln(2^{n}ln(2))^{a}}$ and $ln(2) rightarrow ln(2) $, więc we have to deal only $(ln(2^{n}ln(2))^{a}$, but I still do not know how to do it
                      $endgroup$
                      – MP3129
                      Dec 13 '18 at 20:10










                    • $begingroup$
                      @MP3129 You are almost done indeed $$frac{1}{ln(2)cdot(ln(2^{n}ln(2))^{a}} =frac{1}{ln(2)cdot(ln(2^{n})+ln(ln(2)) )^{a}} =frac{1}{ln(2)cdot n^a(ln(2)+ln(ln(2))/n )^{a}}$$
                      $endgroup$
                      – gimusi
                      Dec 13 '18 at 20:27












                    • $begingroup$
                      But I don't know what I can do with $(ln(2)+ln(ln(2))/n)^{a}$
                      $endgroup$
                      – MP3129
                      Dec 13 '18 at 21:05










                    • $begingroup$
                      @MP3129 Simply note that $(ln(2)+ln(ln(2))/n)^{a}to (ln 2)^a$ then for the series we can refer to limit comparison test with $sum frac1{n^a}$.
                      $endgroup$
                      – gimusi
                      Dec 13 '18 at 21:07


















                    • $begingroup$
                      Ok, so I have a $frac{1}{ln(2)cdot(ln(2^{n}ln(2))^{a}}$ and $ln(2) rightarrow ln(2) $, więc we have to deal only $(ln(2^{n}ln(2))^{a}$, but I still do not know how to do it
                      $endgroup$
                      – MP3129
                      Dec 13 '18 at 20:10










                    • $begingroup$
                      @MP3129 You are almost done indeed $$frac{1}{ln(2)cdot(ln(2^{n}ln(2))^{a}} =frac{1}{ln(2)cdot(ln(2^{n})+ln(ln(2)) )^{a}} =frac{1}{ln(2)cdot n^a(ln(2)+ln(ln(2))/n )^{a}}$$
                      $endgroup$
                      – gimusi
                      Dec 13 '18 at 20:27












                    • $begingroup$
                      But I don't know what I can do with $(ln(2)+ln(ln(2))/n)^{a}$
                      $endgroup$
                      – MP3129
                      Dec 13 '18 at 21:05










                    • $begingroup$
                      @MP3129 Simply note that $(ln(2)+ln(ln(2))/n)^{a}to (ln 2)^a$ then for the series we can refer to limit comparison test with $sum frac1{n^a}$.
                      $endgroup$
                      – gimusi
                      Dec 13 '18 at 21:07
















                    $begingroup$
                    Ok, so I have a $frac{1}{ln(2)cdot(ln(2^{n}ln(2))^{a}}$ and $ln(2) rightarrow ln(2) $, więc we have to deal only $(ln(2^{n}ln(2))^{a}$, but I still do not know how to do it
                    $endgroup$
                    – MP3129
                    Dec 13 '18 at 20:10




                    $begingroup$
                    Ok, so I have a $frac{1}{ln(2)cdot(ln(2^{n}ln(2))^{a}}$ and $ln(2) rightarrow ln(2) $, więc we have to deal only $(ln(2^{n}ln(2))^{a}$, but I still do not know how to do it
                    $endgroup$
                    – MP3129
                    Dec 13 '18 at 20:10












                    $begingroup$
                    @MP3129 You are almost done indeed $$frac{1}{ln(2)cdot(ln(2^{n}ln(2))^{a}} =frac{1}{ln(2)cdot(ln(2^{n})+ln(ln(2)) )^{a}} =frac{1}{ln(2)cdot n^a(ln(2)+ln(ln(2))/n )^{a}}$$
                    $endgroup$
                    – gimusi
                    Dec 13 '18 at 20:27






                    $begingroup$
                    @MP3129 You are almost done indeed $$frac{1}{ln(2)cdot(ln(2^{n}ln(2))^{a}} =frac{1}{ln(2)cdot(ln(2^{n})+ln(ln(2)) )^{a}} =frac{1}{ln(2)cdot n^a(ln(2)+ln(ln(2))/n )^{a}}$$
                    $endgroup$
                    – gimusi
                    Dec 13 '18 at 20:27














                    $begingroup$
                    But I don't know what I can do with $(ln(2)+ln(ln(2))/n)^{a}$
                    $endgroup$
                    – MP3129
                    Dec 13 '18 at 21:05




                    $begingroup$
                    But I don't know what I can do with $(ln(2)+ln(ln(2))/n)^{a}$
                    $endgroup$
                    – MP3129
                    Dec 13 '18 at 21:05












                    $begingroup$
                    @MP3129 Simply note that $(ln(2)+ln(ln(2))/n)^{a}to (ln 2)^a$ then for the series we can refer to limit comparison test with $sum frac1{n^a}$.
                    $endgroup$
                    – gimusi
                    Dec 13 '18 at 21:07




                    $begingroup$
                    @MP3129 Simply note that $(ln(2)+ln(ln(2))/n)^{a}to (ln 2)^a$ then for the series we can refer to limit comparison test with $sum frac1{n^a}$.
                    $endgroup$
                    – gimusi
                    Dec 13 '18 at 21:07


















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