Limit of sequence $lim limits_{n to infty }(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}-frac{sqrt[2018]n}{ln n}$












1












$begingroup$


$lim limits_{n to infty }(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}-frac{sqrt[2018]n}{ln n}$



I can show that $lim limits_{n to infty } {frac{(ln(ln n))^{2019}}{ln n}}=0 $, so $lim limits_{n to infty }(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}=+infty$



and $lim limits_{n to infty } frac{sqrt[2018]n}{ln n}=-infty$



But I don't know how take it together, and show that $lim limits_{n to infty }a_n=-infty$.



Could someone give me hints?






real-analysis sequences-and-series limits limits-without-lhopital







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you familiar with Taylor series (as an available tool)?
    $endgroup$
    – Clement C.
    Dec 10 '18 at 21:19












  • $begingroup$
    No, I can't use it :/
    $endgroup$
    – matematiccc
    Dec 10 '18 at 21:21










  • $begingroup$
    Where does this problem come from? Could you write down the source?
    $endgroup$
    – user376343
    Dec 10 '18 at 22:10
















1












$begingroup$


$lim limits_{n to infty }(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}-frac{sqrt[2018]n}{ln n}$



I can show that $lim limits_{n to infty } {frac{(ln(ln n))^{2019}}{ln n}}=0 $, so $lim limits_{n to infty }(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}=+infty$



and $lim limits_{n to infty } frac{sqrt[2018]n}{ln n}=-infty$



But I don't know how take it together, and show that $lim limits_{n to infty }a_n=-infty$.



Could someone give me hints?






real-analysis sequences-and-series limits limits-without-lhopital







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you familiar with Taylor series (as an available tool)?
    $endgroup$
    – Clement C.
    Dec 10 '18 at 21:19












  • $begingroup$
    No, I can't use it :/
    $endgroup$
    – matematiccc
    Dec 10 '18 at 21:21










  • $begingroup$
    Where does this problem come from? Could you write down the source?
    $endgroup$
    – user376343
    Dec 10 '18 at 22:10














1












1








1


1



$begingroup$


$lim limits_{n to infty }(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}-frac{sqrt[2018]n}{ln n}$



I can show that $lim limits_{n to infty } {frac{(ln(ln n))^{2019}}{ln n}}=0 $, so $lim limits_{n to infty }(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}=+infty$



and $lim limits_{n to infty } frac{sqrt[2018]n}{ln n}=-infty$



But I don't know how take it together, and show that $lim limits_{n to infty }a_n=-infty$.



Could someone give me hints?






real-analysis sequences-and-series limits limits-without-lhopital







share|cite|improve this question











$endgroup$




$lim limits_{n to infty }(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}-frac{sqrt[2018]n}{ln n}$



I can show that $lim limits_{n to infty } {frac{(ln(ln n))^{2019}}{ln n}}=0 $, so $lim limits_{n to infty }(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}=+infty$



and $lim limits_{n to infty } frac{sqrt[2018]n}{ln n}=-infty$



But I don't know how take it together, and show that $lim limits_{n to infty }a_n=-infty$.



Could someone give me hints?






real-analysis sequences-and-series limits limits-without-lhopital




real-analysis sequences-and-series limits limits-without-lhopital






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 21:11









user376343

3,7383827




3,7383827










asked Dec 10 '18 at 21:05









matematicccmatematiccc

1275




1275












  • $begingroup$
    Are you familiar with Taylor series (as an available tool)?
    $endgroup$
    – Clement C.
    Dec 10 '18 at 21:19












  • $begingroup$
    No, I can't use it :/
    $endgroup$
    – matematiccc
    Dec 10 '18 at 21:21










  • $begingroup$
    Where does this problem come from? Could you write down the source?
    $endgroup$
    – user376343
    Dec 10 '18 at 22:10


















  • $begingroup$
    Are you familiar with Taylor series (as an available tool)?
    $endgroup$
    – Clement C.
    Dec 10 '18 at 21:19












  • $begingroup$
    No, I can't use it :/
    $endgroup$
    – matematiccc
    Dec 10 '18 at 21:21










  • $begingroup$
    Where does this problem come from? Could you write down the source?
    $endgroup$
    – user376343
    Dec 10 '18 at 22:10
















$begingroup$
Are you familiar with Taylor series (as an available tool)?
$endgroup$
– Clement C.
Dec 10 '18 at 21:19






$begingroup$
Are you familiar with Taylor series (as an available tool)?
$endgroup$
– Clement C.
Dec 10 '18 at 21:19














$begingroup$
No, I can't use it :/
$endgroup$
– matematiccc
Dec 10 '18 at 21:21




$begingroup$
No, I can't use it :/
$endgroup$
– matematiccc
Dec 10 '18 at 21:21












$begingroup$
Where does this problem come from? Could you write down the source?
$endgroup$
– user376343
Dec 10 '18 at 22:10




$begingroup$
Where does this problem come from? Could you write down the source?
$endgroup$
– user376343
Dec 10 '18 at 22:10










1 Answer
1






active

oldest

votes


















3












$begingroup$

We have that



$$(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}=ln n cdot e^{frac{(ln(ln n))^{2020}}{ln n}}$$



and then



$$(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}-frac{sqrt[2018]n}{ln n}= ln n cdot e^{frac{(ln(ln n))^{2020}}{ln n}}-frac{sqrt[2018]n}{ln n}=ln n left(e^{frac{(ln(ln n))^{2020}}{ln n}}-frac{sqrt[2018]n}{ln^2 n}right)to-infty$$



indeed $forall a>0$




  • $frac{n^a}{log n} to infty$

  • $frac{(ln(ln n))^{a}}{ln n} to 0$






share|cite|improve this answer









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    3












    $begingroup$

    We have that



    $$(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}=ln n cdot e^{frac{(ln(ln n))^{2020}}{ln n}}$$



    and then



    $$(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}-frac{sqrt[2018]n}{ln n}= ln n cdot e^{frac{(ln(ln n))^{2020}}{ln n}}-frac{sqrt[2018]n}{ln n}=ln n left(e^{frac{(ln(ln n))^{2020}}{ln n}}-frac{sqrt[2018]n}{ln^2 n}right)to-infty$$



    indeed $forall a>0$




    • $frac{n^a}{log n} to infty$

    • $frac{(ln(ln n))^{a}}{ln n} to 0$






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      We have that



      $$(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}=ln n cdot e^{frac{(ln(ln n))^{2020}}{ln n}}$$



      and then



      $$(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}-frac{sqrt[2018]n}{ln n}= ln n cdot e^{frac{(ln(ln n))^{2020}}{ln n}}-frac{sqrt[2018]n}{ln n}=ln n left(e^{frac{(ln(ln n))^{2020}}{ln n}}-frac{sqrt[2018]n}{ln^2 n}right)to-infty$$



      indeed $forall a>0$




      • $frac{n^a}{log n} to infty$

      • $frac{(ln(ln n))^{a}}{ln n} to 0$






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        We have that



        $$(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}=ln n cdot e^{frac{(ln(ln n))^{2020}}{ln n}}$$



        and then



        $$(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}-frac{sqrt[2018]n}{ln n}= ln n cdot e^{frac{(ln(ln n))^{2020}}{ln n}}-frac{sqrt[2018]n}{ln n}=ln n left(e^{frac{(ln(ln n))^{2020}}{ln n}}-frac{sqrt[2018]n}{ln^2 n}right)to-infty$$



        indeed $forall a>0$




        • $frac{n^a}{log n} to infty$

        • $frac{(ln(ln n))^{a}}{ln n} to 0$






        share|cite|improve this answer









        $endgroup$



        We have that



        $$(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}=ln n cdot e^{frac{(ln(ln n))^{2020}}{ln n}}$$



        and then



        $$(ln n)^{1+frac{(ln(ln n))^{2019}}{ln n}}-frac{sqrt[2018]n}{ln n}= ln n cdot e^{frac{(ln(ln n))^{2020}}{ln n}}-frac{sqrt[2018]n}{ln n}=ln n left(e^{frac{(ln(ln n))^{2020}}{ln n}}-frac{sqrt[2018]n}{ln^2 n}right)to-infty$$



        indeed $forall a>0$




        • $frac{n^a}{log n} to infty$

        • $frac{(ln(ln n))^{a}}{ln n} to 0$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 22:27









        gimusigimusi

        92.8k84494




        92.8k84494






























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