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






      share|cite|improve this question















      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

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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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                    Ida-Boy-Ed-Garten

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                    Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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