$lim_{n to infty} dfrac{e^{c sqrt{ln n . ln ln n}}}{n^{epsilon}}=0$?












1












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The general rule is discussed here but that doesn't solve my problem. I want to prove that $$lim_{n to infty} dfrac{e^{c sqrt{ln n . ln ln n}}}{n^{epsilon}}=0$$ where $c>0$ a fixed constant and $epsilon>0$ is any small positive number. I used different methods like l'hopital, etc and I found by expanding $e^{f(n)}$ it is the easiest, but need to know :



$lim_{n to infty} dfrac{(ln n . ln ln n)^k}{n^{epsilon}}=0$ for any $k in mathbb{N}$. Does it imply $lim_{n to infty} dfrac{e^{c sqrt{ln n . ln ln n}}}{n^{epsilon}}=0$?



If not, any hint for solving $lim_{n to infty} dfrac{e^{c sqrt{ln n . ln ln n}}}{n^{epsilon}}=0$ would be appreciated.










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    $begingroup$


    The general rule is discussed here but that doesn't solve my problem. I want to prove that $$lim_{n to infty} dfrac{e^{c sqrt{ln n . ln ln n}}}{n^{epsilon}}=0$$ where $c>0$ a fixed constant and $epsilon>0$ is any small positive number. I used different methods like l'hopital, etc and I found by expanding $e^{f(n)}$ it is the easiest, but need to know :



    $lim_{n to infty} dfrac{(ln n . ln ln n)^k}{n^{epsilon}}=0$ for any $k in mathbb{N}$. Does it imply $lim_{n to infty} dfrac{e^{c sqrt{ln n . ln ln n}}}{n^{epsilon}}=0$?



    If not, any hint for solving $lim_{n to infty} dfrac{e^{c sqrt{ln n . ln ln n}}}{n^{epsilon}}=0$ would be appreciated.










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      The general rule is discussed here but that doesn't solve my problem. I want to prove that $$lim_{n to infty} dfrac{e^{c sqrt{ln n . ln ln n}}}{n^{epsilon}}=0$$ where $c>0$ a fixed constant and $epsilon>0$ is any small positive number. I used different methods like l'hopital, etc and I found by expanding $e^{f(n)}$ it is the easiest, but need to know :



      $lim_{n to infty} dfrac{(ln n . ln ln n)^k}{n^{epsilon}}=0$ for any $k in mathbb{N}$. Does it imply $lim_{n to infty} dfrac{e^{c sqrt{ln n . ln ln n}}}{n^{epsilon}}=0$?



      If not, any hint for solving $lim_{n to infty} dfrac{e^{c sqrt{ln n . ln ln n}}}{n^{epsilon}}=0$ would be appreciated.










      share|cite|improve this question









      $endgroup$




      The general rule is discussed here but that doesn't solve my problem. I want to prove that $$lim_{n to infty} dfrac{e^{c sqrt{ln n . ln ln n}}}{n^{epsilon}}=0$$ where $c>0$ a fixed constant and $epsilon>0$ is any small positive number. I used different methods like l'hopital, etc and I found by expanding $e^{f(n)}$ it is the easiest, but need to know :



      $lim_{n to infty} dfrac{(ln n . ln ln n)^k}{n^{epsilon}}=0$ for any $k in mathbb{N}$. Does it imply $lim_{n to infty} dfrac{e^{c sqrt{ln n . ln ln n}}}{n^{epsilon}}=0$?



      If not, any hint for solving $lim_{n to infty} dfrac{e^{c sqrt{ln n . ln ln n}}}{n^{epsilon}}=0$ would be appreciated.







      limits analytic-number-theory






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      asked Dec 21 '18 at 10:02









      72D72D

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          2 Answers
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          $begingroup$

          $$displaystyle L=lim_{ntoinfty}frac{e^{csqrt{ln ncdotlnln n}}}{n^{epsilon}}$$



          $displaystyleln L=lim_{ntoinfty}csqrt{ln ncdotlnln n}-epsilonln n$



          $displaystyle=lim_{xtoinfty}csqrt{xcdotln x}-epsilon x$



          $displaystyle=lim_{xtoinfty}frac{csqrt{frac{ln x}x}-epsilon}{frac1x}to-infty becauselim_{xtoinfty}frac{ln x}x=0 $



          $therefore Lto e^{-infty}=0$






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            $begingroup$

            The statement is equivalent to $csqrt {ln, n lnln , n}-epsilon ln, n to -infty$. To show this write this as $-ln, n [frac {csqrt {ln, n lnln , n}} {-ln , n} +epsilon]$. Can you show that the expression inside goes to $epsilon$? It amounts to showing that $frac {ln ln ,n} {ln, n} to 0$ for which you can use L'Hopital's Rule.






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              2 Answers
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              2 Answers
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              1












              $begingroup$

              $$displaystyle L=lim_{ntoinfty}frac{e^{csqrt{ln ncdotlnln n}}}{n^{epsilon}}$$



              $displaystyleln L=lim_{ntoinfty}csqrt{ln ncdotlnln n}-epsilonln n$



              $displaystyle=lim_{xtoinfty}csqrt{xcdotln x}-epsilon x$



              $displaystyle=lim_{xtoinfty}frac{csqrt{frac{ln x}x}-epsilon}{frac1x}to-infty becauselim_{xtoinfty}frac{ln x}x=0 $



              $therefore Lto e^{-infty}=0$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                $$displaystyle L=lim_{ntoinfty}frac{e^{csqrt{ln ncdotlnln n}}}{n^{epsilon}}$$



                $displaystyleln L=lim_{ntoinfty}csqrt{ln ncdotlnln n}-epsilonln n$



                $displaystyle=lim_{xtoinfty}csqrt{xcdotln x}-epsilon x$



                $displaystyle=lim_{xtoinfty}frac{csqrt{frac{ln x}x}-epsilon}{frac1x}to-infty becauselim_{xtoinfty}frac{ln x}x=0 $



                $therefore Lto e^{-infty}=0$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $$displaystyle L=lim_{ntoinfty}frac{e^{csqrt{ln ncdotlnln n}}}{n^{epsilon}}$$



                  $displaystyleln L=lim_{ntoinfty}csqrt{ln ncdotlnln n}-epsilonln n$



                  $displaystyle=lim_{xtoinfty}csqrt{xcdotln x}-epsilon x$



                  $displaystyle=lim_{xtoinfty}frac{csqrt{frac{ln x}x}-epsilon}{frac1x}to-infty becauselim_{xtoinfty}frac{ln x}x=0 $



                  $therefore Lto e^{-infty}=0$






                  share|cite|improve this answer









                  $endgroup$



                  $$displaystyle L=lim_{ntoinfty}frac{e^{csqrt{ln ncdotlnln n}}}{n^{epsilon}}$$



                  $displaystyleln L=lim_{ntoinfty}csqrt{ln ncdotlnln n}-epsilonln n$



                  $displaystyle=lim_{xtoinfty}csqrt{xcdotln x}-epsilon x$



                  $displaystyle=lim_{xtoinfty}frac{csqrt{frac{ln x}x}-epsilon}{frac1x}to-infty becauselim_{xtoinfty}frac{ln x}x=0 $



                  $therefore Lto e^{-infty}=0$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 21 '18 at 10:16









                  Shubham JohriShubham Johri

                  5,192717




                  5,192717























                      1












                      $begingroup$

                      The statement is equivalent to $csqrt {ln, n lnln , n}-epsilon ln, n to -infty$. To show this write this as $-ln, n [frac {csqrt {ln, n lnln , n}} {-ln , n} +epsilon]$. Can you show that the expression inside goes to $epsilon$? It amounts to showing that $frac {ln ln ,n} {ln, n} to 0$ for which you can use L'Hopital's Rule.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        The statement is equivalent to $csqrt {ln, n lnln , n}-epsilon ln, n to -infty$. To show this write this as $-ln, n [frac {csqrt {ln, n lnln , n}} {-ln , n} +epsilon]$. Can you show that the expression inside goes to $epsilon$? It amounts to showing that $frac {ln ln ,n} {ln, n} to 0$ for which you can use L'Hopital's Rule.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          The statement is equivalent to $csqrt {ln, n lnln , n}-epsilon ln, n to -infty$. To show this write this as $-ln, n [frac {csqrt {ln, n lnln , n}} {-ln , n} +epsilon]$. Can you show that the expression inside goes to $epsilon$? It amounts to showing that $frac {ln ln ,n} {ln, n} to 0$ for which you can use L'Hopital's Rule.






                          share|cite|improve this answer









                          $endgroup$



                          The statement is equivalent to $csqrt {ln, n lnln , n}-epsilon ln, n to -infty$. To show this write this as $-ln, n [frac {csqrt {ln, n lnln , n}} {-ln , n} +epsilon]$. Can you show that the expression inside goes to $epsilon$? It amounts to showing that $frac {ln ln ,n} {ln, n} to 0$ for which you can use L'Hopital's Rule.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 21 '18 at 10:14









                          Kavi Rama MurthyKavi Rama Murthy

                          60.6k42161




                          60.6k42161






























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