Show that the following limit is infinity












0












$begingroup$


I want to show that the limit of $$frac{1}{n^alpha cdot ln(n)}$$ is infinity where $alpha < 0$. Is there an elementary way to show this? I know that $n^alpha$ gets small faster then $ln(n)$ gets large but still I can't find a formal proof. Thanks!



Edit: I look for a solution that uses only sequences rules and theorems, since I did not learn about functions yet.






limits







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$endgroup$












  • $begingroup$
    You are looking for a solution only by sequence? You should add that to your question in order to make it clear.
    $endgroup$
    – gimusi
    Dec 9 '18 at 15:45










  • $begingroup$
    @gimusi Okay. sorry. I edited the post.
    $endgroup$
    – Omer
    Dec 9 '18 at 15:48






  • 2




    $begingroup$
    What properties of the $log$ function do you know, yet?
    $endgroup$
    – Martin Rosenau
    Dec 9 '18 at 16:13
















0












$begingroup$


I want to show that the limit of $$frac{1}{n^alpha cdot ln(n)}$$ is infinity where $alpha < 0$. Is there an elementary way to show this? I know that $n^alpha$ gets small faster then $ln(n)$ gets large but still I can't find a formal proof. Thanks!



Edit: I look for a solution that uses only sequences rules and theorems, since I did not learn about functions yet.






limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    You are looking for a solution only by sequence? You should add that to your question in order to make it clear.
    $endgroup$
    – gimusi
    Dec 9 '18 at 15:45










  • $begingroup$
    @gimusi Okay. sorry. I edited the post.
    $endgroup$
    – Omer
    Dec 9 '18 at 15:48






  • 2




    $begingroup$
    What properties of the $log$ function do you know, yet?
    $endgroup$
    – Martin Rosenau
    Dec 9 '18 at 16:13














0












0








0


1



$begingroup$


I want to show that the limit of $$frac{1}{n^alpha cdot ln(n)}$$ is infinity where $alpha < 0$. Is there an elementary way to show this? I know that $n^alpha$ gets small faster then $ln(n)$ gets large but still I can't find a formal proof. Thanks!



Edit: I look for a solution that uses only sequences rules and theorems, since I did not learn about functions yet.






limits







share|cite|improve this question











$endgroup$




I want to show that the limit of $$frac{1}{n^alpha cdot ln(n)}$$ is infinity where $alpha < 0$. Is there an elementary way to show this? I know that $n^alpha$ gets small faster then $ln(n)$ gets large but still I can't find a formal proof. Thanks!



Edit: I look for a solution that uses only sequences rules and theorems, since I did not learn about functions yet.






limits




limits






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 16:48









Namaste

1




1










asked Dec 9 '18 at 15:32









OmerOmer

4039




4039












  • $begingroup$
    You are looking for a solution only by sequence? You should add that to your question in order to make it clear.
    $endgroup$
    – gimusi
    Dec 9 '18 at 15:45










  • $begingroup$
    @gimusi Okay. sorry. I edited the post.
    $endgroup$
    – Omer
    Dec 9 '18 at 15:48






  • 2




    $begingroup$
    What properties of the $log$ function do you know, yet?
    $endgroup$
    – Martin Rosenau
    Dec 9 '18 at 16:13


















  • $begingroup$
    You are looking for a solution only by sequence? You should add that to your question in order to make it clear.
    $endgroup$
    – gimusi
    Dec 9 '18 at 15:45










  • $begingroup$
    @gimusi Okay. sorry. I edited the post.
    $endgroup$
    – Omer
    Dec 9 '18 at 15:48






  • 2




    $begingroup$
    What properties of the $log$ function do you know, yet?
    $endgroup$
    – Martin Rosenau
    Dec 9 '18 at 16:13
















$begingroup$
You are looking for a solution only by sequence? You should add that to your question in order to make it clear.
$endgroup$
– gimusi
Dec 9 '18 at 15:45




$begingroup$
You are looking for a solution only by sequence? You should add that to your question in order to make it clear.
$endgroup$
– gimusi
Dec 9 '18 at 15:45












$begingroup$
@gimusi Okay. sorry. I edited the post.
$endgroup$
– Omer
Dec 9 '18 at 15:48




$begingroup$
@gimusi Okay. sorry. I edited the post.
$endgroup$
– Omer
Dec 9 '18 at 15:48




2




2




$begingroup$
What properties of the $log$ function do you know, yet?
$endgroup$
– Martin Rosenau
Dec 9 '18 at 16:13




$begingroup$
What properties of the $log$ function do you know, yet?
$endgroup$
– Martin Rosenau
Dec 9 '18 at 16:13










4 Answers
4






active

oldest

votes


















2












$begingroup$

A classic proof in high school that $;limdfrac{log x}x=0$ consists in observing that, since $tgesqrt t$ for $tge 1$, we have
$$log x=int_1^xfrac{mathrm dt}tle int_1^xfrac{mathrm dt}{sqrt t}=2bigl(sqrt x-1bigr)<2sqrt x,$$
hence $;dfrac{log x}x<dfrac{2sqrt x}x=dfrac 2{sqrt x}$, which tends to $0$ as $x$ tends to $infty$.



Now , setting $alpha =-a;(a>0)$, we deduce that
$$frac{1}{n^alpha cdot log(n)}=frac{n^a}{log n}=afrac{n^a}{log(n^a)}longrightarrow acdot +infty=+infty$$
by the above result.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    "I look for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
    $endgroup$
    – gimusi
    Dec 9 '18 at 16:01










  • $begingroup$
    How does the O.P. know about the log, then?
    $endgroup$
    – Bernard
    Dec 9 '18 at 16:02










  • $begingroup$
    @gimusi: I don't see whay, in your 2nd displayed equation, the limit is $e^b$. Where has $a$ gone?
    $endgroup$
    – Bernard
    Dec 9 '18 at 16:51










  • $begingroup$
    It seems you miscalculated $;sqrt[n]{e^{n^a}}$: it is equal to $;e^{n^{a-1}}$.
    $endgroup$
    – Bernard
    Dec 9 '18 at 17:03










  • $begingroup$
    What about the new version in your opinion?
    $endgroup$
    – gimusi
    Dec 9 '18 at 18:16



















1












$begingroup$

Attempt:



$a>0$.



$dfrac{n^a}{log n}= dfrac{exp(log n^a)}{log n}=$



$dfrac{exp (alog n)}{log n}.$



$z:= log n.$



Now consider $z rightarrow infty$.



$f(z)=dfrac{exp (az)}{z} gt (a^2/2!)z.$



Hence $lim_{z rightarrow infty} f(z)=infty.$



Used:



$e^x =1+x +x^2/2!...> x^2/2!$, $x >0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That seems nice even if the OP is looking "for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
    $endgroup$
    – gimusi
    Dec 9 '18 at 18:16










  • $begingroup$
    Gimusi.True.Delete?Your thoughts?
    $endgroup$
    – Peter Szilas
    Dec 9 '18 at 18:30










  • $begingroup$
    My opinion is that you can let it of course. There are also other answers with more advanced approach than what requested by the asker, and your one is different. Moreover some other reader could be interested to it.
    $endgroup$
    – gimusi
    Dec 9 '18 at 18:34



















0












$begingroup$

With
$$ln n<n-1$$
then
$$dfrac{1}{n^aln n}>dfrac{1}{n-1}n^{-a}toinfty$$
for $-a>1$.




Edit:
Might be helpful:
$$ln x=int_1^xfrac{mathrm dt}tle int_1^xfrac{t^bmathrm dt}{t}=int_1^x t^{b-1}mathrm dt=frac{x^b-1}{b}<frac{x^b}{b}$$
for real $b>0$. Now let $b=dfrac{a}{2}$, then
$$dfrac{1}{n^aln n}>dfrac{1}{n^frac{a}{2}n^frac{a}{2}}toinfty$$
if $-a>0$.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    You don't have the case $--1le a<0$.
    $endgroup$
    – Bernard
    Dec 9 '18 at 16:01










  • $begingroup$
    the rest is proved with integrals and isn't elementary.
    $endgroup$
    – Nosrati
    Dec 9 '18 at 16:02










  • $begingroup$
    For me it's elementary inasmuch as it only uses elementary properties of integrals, as seen in high school. Or the O.P. should give us the definition of the log he/she uses.
    $endgroup$
    – Bernard
    Dec 9 '18 at 16:06










  • $begingroup$
    Sorry fro the (maybe stupid) question, but the first inequality is assumed as a given?
    $endgroup$
    – gimusi
    Dec 9 '18 at 19:19










  • $begingroup$
    As you know that one is proven with calculus, Isn't that?
    $endgroup$
    – Nosrati
    Dec 9 '18 at 19:21





















-2












$begingroup$

Preliminary results



We have that $forall a>0$



$$frac{e^{n}} {n^a} to infty$$



indeed by ratio test



$$frac{e^{n+1}}{(n+1)^a}frac {n^a} {e^n}to e>1$$



then from here we have that



$$frac{log n}{n} to 0 iff frac n {log n}to infty$$



indeed for any $a >0$



$$frac n {log n}ge a iff nge log n^a iff e^nge n^a$$



Proof of the main result



By $b=-alpha>0$ we have



$$frac{1}{n^alpha cdot ln n }=frac{n^b}{ln n}=bfrac{n^b}{bln n}=bfrac{n^b}{ln n^b}$$



and since for any $b>0 implies n^bto infty$ we have that



$$bfrac{n^b}{ln n^b}to infty implies frac{1}{n^alpha cdot ln n } to infty$$






share|cite|improve this answer











$endgroup$













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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    A classic proof in high school that $;limdfrac{log x}x=0$ consists in observing that, since $tgesqrt t$ for $tge 1$, we have
    $$log x=int_1^xfrac{mathrm dt}tle int_1^xfrac{mathrm dt}{sqrt t}=2bigl(sqrt x-1bigr)<2sqrt x,$$
    hence $;dfrac{log x}x<dfrac{2sqrt x}x=dfrac 2{sqrt x}$, which tends to $0$ as $x$ tends to $infty$.



    Now , setting $alpha =-a;(a>0)$, we deduce that
    $$frac{1}{n^alpha cdot log(n)}=frac{n^a}{log n}=afrac{n^a}{log(n^a)}longrightarrow acdot +infty=+infty$$
    by the above result.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      "I look for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
      $endgroup$
      – gimusi
      Dec 9 '18 at 16:01










    • $begingroup$
      How does the O.P. know about the log, then?
      $endgroup$
      – Bernard
      Dec 9 '18 at 16:02










    • $begingroup$
      @gimusi: I don't see whay, in your 2nd displayed equation, the limit is $e^b$. Where has $a$ gone?
      $endgroup$
      – Bernard
      Dec 9 '18 at 16:51










    • $begingroup$
      It seems you miscalculated $;sqrt[n]{e^{n^a}}$: it is equal to $;e^{n^{a-1}}$.
      $endgroup$
      – Bernard
      Dec 9 '18 at 17:03










    • $begingroup$
      What about the new version in your opinion?
      $endgroup$
      – gimusi
      Dec 9 '18 at 18:16
















    2












    $begingroup$

    A classic proof in high school that $;limdfrac{log x}x=0$ consists in observing that, since $tgesqrt t$ for $tge 1$, we have
    $$log x=int_1^xfrac{mathrm dt}tle int_1^xfrac{mathrm dt}{sqrt t}=2bigl(sqrt x-1bigr)<2sqrt x,$$
    hence $;dfrac{log x}x<dfrac{2sqrt x}x=dfrac 2{sqrt x}$, which tends to $0$ as $x$ tends to $infty$.



    Now , setting $alpha =-a;(a>0)$, we deduce that
    $$frac{1}{n^alpha cdot log(n)}=frac{n^a}{log n}=afrac{n^a}{log(n^a)}longrightarrow acdot +infty=+infty$$
    by the above result.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      "I look for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
      $endgroup$
      – gimusi
      Dec 9 '18 at 16:01










    • $begingroup$
      How does the O.P. know about the log, then?
      $endgroup$
      – Bernard
      Dec 9 '18 at 16:02










    • $begingroup$
      @gimusi: I don't see whay, in your 2nd displayed equation, the limit is $e^b$. Where has $a$ gone?
      $endgroup$
      – Bernard
      Dec 9 '18 at 16:51










    • $begingroup$
      It seems you miscalculated $;sqrt[n]{e^{n^a}}$: it is equal to $;e^{n^{a-1}}$.
      $endgroup$
      – Bernard
      Dec 9 '18 at 17:03










    • $begingroup$
      What about the new version in your opinion?
      $endgroup$
      – gimusi
      Dec 9 '18 at 18:16














    2












    2








    2





    $begingroup$

    A classic proof in high school that $;limdfrac{log x}x=0$ consists in observing that, since $tgesqrt t$ for $tge 1$, we have
    $$log x=int_1^xfrac{mathrm dt}tle int_1^xfrac{mathrm dt}{sqrt t}=2bigl(sqrt x-1bigr)<2sqrt x,$$
    hence $;dfrac{log x}x<dfrac{2sqrt x}x=dfrac 2{sqrt x}$, which tends to $0$ as $x$ tends to $infty$.



    Now , setting $alpha =-a;(a>0)$, we deduce that
    $$frac{1}{n^alpha cdot log(n)}=frac{n^a}{log n}=afrac{n^a}{log(n^a)}longrightarrow acdot +infty=+infty$$
    by the above result.






    share|cite|improve this answer











    $endgroup$



    A classic proof in high school that $;limdfrac{log x}x=0$ consists in observing that, since $tgesqrt t$ for $tge 1$, we have
    $$log x=int_1^xfrac{mathrm dt}tle int_1^xfrac{mathrm dt}{sqrt t}=2bigl(sqrt x-1bigr)<2sqrt x,$$
    hence $;dfrac{log x}x<dfrac{2sqrt x}x=dfrac 2{sqrt x}$, which tends to $0$ as $x$ tends to $infty$.



    Now , setting $alpha =-a;(a>0)$, we deduce that
    $$frac{1}{n^alpha cdot log(n)}=frac{n^a}{log n}=afrac{n^a}{log(n^a)}longrightarrow acdot +infty=+infty$$
    by the above result.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 9 '18 at 19:15

























    answered Dec 9 '18 at 15:59









    BernardBernard

    123k741117




    123k741117












    • $begingroup$
      "I look for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
      $endgroup$
      – gimusi
      Dec 9 '18 at 16:01










    • $begingroup$
      How does the O.P. know about the log, then?
      $endgroup$
      – Bernard
      Dec 9 '18 at 16:02










    • $begingroup$
      @gimusi: I don't see whay, in your 2nd displayed equation, the limit is $e^b$. Where has $a$ gone?
      $endgroup$
      – Bernard
      Dec 9 '18 at 16:51










    • $begingroup$
      It seems you miscalculated $;sqrt[n]{e^{n^a}}$: it is equal to $;e^{n^{a-1}}$.
      $endgroup$
      – Bernard
      Dec 9 '18 at 17:03










    • $begingroup$
      What about the new version in your opinion?
      $endgroup$
      – gimusi
      Dec 9 '18 at 18:16


















    • $begingroup$
      "I look for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
      $endgroup$
      – gimusi
      Dec 9 '18 at 16:01










    • $begingroup$
      How does the O.P. know about the log, then?
      $endgroup$
      – Bernard
      Dec 9 '18 at 16:02










    • $begingroup$
      @gimusi: I don't see whay, in your 2nd displayed equation, the limit is $e^b$. Where has $a$ gone?
      $endgroup$
      – Bernard
      Dec 9 '18 at 16:51










    • $begingroup$
      It seems you miscalculated $;sqrt[n]{e^{n^a}}$: it is equal to $;e^{n^{a-1}}$.
      $endgroup$
      – Bernard
      Dec 9 '18 at 17:03










    • $begingroup$
      What about the new version in your opinion?
      $endgroup$
      – gimusi
      Dec 9 '18 at 18:16
















    $begingroup$
    "I look for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
    $endgroup$
    – gimusi
    Dec 9 '18 at 16:01




    $begingroup$
    "I look for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
    $endgroup$
    – gimusi
    Dec 9 '18 at 16:01












    $begingroup$
    How does the O.P. know about the log, then?
    $endgroup$
    – Bernard
    Dec 9 '18 at 16:02




    $begingroup$
    How does the O.P. know about the log, then?
    $endgroup$
    – Bernard
    Dec 9 '18 at 16:02












    $begingroup$
    @gimusi: I don't see whay, in your 2nd displayed equation, the limit is $e^b$. Where has $a$ gone?
    $endgroup$
    – Bernard
    Dec 9 '18 at 16:51




    $begingroup$
    @gimusi: I don't see whay, in your 2nd displayed equation, the limit is $e^b$. Where has $a$ gone?
    $endgroup$
    – Bernard
    Dec 9 '18 at 16:51












    $begingroup$
    It seems you miscalculated $;sqrt[n]{e^{n^a}}$: it is equal to $;e^{n^{a-1}}$.
    $endgroup$
    – Bernard
    Dec 9 '18 at 17:03




    $begingroup$
    It seems you miscalculated $;sqrt[n]{e^{n^a}}$: it is equal to $;e^{n^{a-1}}$.
    $endgroup$
    – Bernard
    Dec 9 '18 at 17:03












    $begingroup$
    What about the new version in your opinion?
    $endgroup$
    – gimusi
    Dec 9 '18 at 18:16




    $begingroup$
    What about the new version in your opinion?
    $endgroup$
    – gimusi
    Dec 9 '18 at 18:16











    1












    $begingroup$

    Attempt:



    $a>0$.



    $dfrac{n^a}{log n}= dfrac{exp(log n^a)}{log n}=$



    $dfrac{exp (alog n)}{log n}.$



    $z:= log n.$



    Now consider $z rightarrow infty$.



    $f(z)=dfrac{exp (az)}{z} gt (a^2/2!)z.$



    Hence $lim_{z rightarrow infty} f(z)=infty.$



    Used:



    $e^x =1+x +x^2/2!...> x^2/2!$, $x >0$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      That seems nice even if the OP is looking "for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
      $endgroup$
      – gimusi
      Dec 9 '18 at 18:16










    • $begingroup$
      Gimusi.True.Delete?Your thoughts?
      $endgroup$
      – Peter Szilas
      Dec 9 '18 at 18:30










    • $begingroup$
      My opinion is that you can let it of course. There are also other answers with more advanced approach than what requested by the asker, and your one is different. Moreover some other reader could be interested to it.
      $endgroup$
      – gimusi
      Dec 9 '18 at 18:34
















    1












    $begingroup$

    Attempt:



    $a>0$.



    $dfrac{n^a}{log n}= dfrac{exp(log n^a)}{log n}=$



    $dfrac{exp (alog n)}{log n}.$



    $z:= log n.$



    Now consider $z rightarrow infty$.



    $f(z)=dfrac{exp (az)}{z} gt (a^2/2!)z.$



    Hence $lim_{z rightarrow infty} f(z)=infty.$



    Used:



    $e^x =1+x +x^2/2!...> x^2/2!$, $x >0$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      That seems nice even if the OP is looking "for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
      $endgroup$
      – gimusi
      Dec 9 '18 at 18:16










    • $begingroup$
      Gimusi.True.Delete?Your thoughts?
      $endgroup$
      – Peter Szilas
      Dec 9 '18 at 18:30










    • $begingroup$
      My opinion is that you can let it of course. There are also other answers with more advanced approach than what requested by the asker, and your one is different. Moreover some other reader could be interested to it.
      $endgroup$
      – gimusi
      Dec 9 '18 at 18:34














    1












    1








    1





    $begingroup$

    Attempt:



    $a>0$.



    $dfrac{n^a}{log n}= dfrac{exp(log n^a)}{log n}=$



    $dfrac{exp (alog n)}{log n}.$



    $z:= log n.$



    Now consider $z rightarrow infty$.



    $f(z)=dfrac{exp (az)}{z} gt (a^2/2!)z.$



    Hence $lim_{z rightarrow infty} f(z)=infty.$



    Used:



    $e^x =1+x +x^2/2!...> x^2/2!$, $x >0$.






    share|cite|improve this answer









    $endgroup$



    Attempt:



    $a>0$.



    $dfrac{n^a}{log n}= dfrac{exp(log n^a)}{log n}=$



    $dfrac{exp (alog n)}{log n}.$



    $z:= log n.$



    Now consider $z rightarrow infty$.



    $f(z)=dfrac{exp (az)}{z} gt (a^2/2!)z.$



    Hence $lim_{z rightarrow infty} f(z)=infty.$



    Used:



    $e^x =1+x +x^2/2!...> x^2/2!$, $x >0$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 9 '18 at 17:46









    Peter SzilasPeter Szilas

    11.6k2822




    11.6k2822












    • $begingroup$
      That seems nice even if the OP is looking "for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
      $endgroup$
      – gimusi
      Dec 9 '18 at 18:16










    • $begingroup$
      Gimusi.True.Delete?Your thoughts?
      $endgroup$
      – Peter Szilas
      Dec 9 '18 at 18:30










    • $begingroup$
      My opinion is that you can let it of course. There are also other answers with more advanced approach than what requested by the asker, and your one is different. Moreover some other reader could be interested to it.
      $endgroup$
      – gimusi
      Dec 9 '18 at 18:34


















    • $begingroup$
      That seems nice even if the OP is looking "for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
      $endgroup$
      – gimusi
      Dec 9 '18 at 18:16










    • $begingroup$
      Gimusi.True.Delete?Your thoughts?
      $endgroup$
      – Peter Szilas
      Dec 9 '18 at 18:30










    • $begingroup$
      My opinion is that you can let it of course. There are also other answers with more advanced approach than what requested by the asker, and your one is different. Moreover some other reader could be interested to it.
      $endgroup$
      – gimusi
      Dec 9 '18 at 18:34
















    $begingroup$
    That seems nice even if the OP is looking "for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
    $endgroup$
    – gimusi
    Dec 9 '18 at 18:16




    $begingroup$
    That seems nice even if the OP is looking "for a solution that uses only sequences rules and theorems, since I did not learn about functions yet"
    $endgroup$
    – gimusi
    Dec 9 '18 at 18:16












    $begingroup$
    Gimusi.True.Delete?Your thoughts?
    $endgroup$
    – Peter Szilas
    Dec 9 '18 at 18:30




    $begingroup$
    Gimusi.True.Delete?Your thoughts?
    $endgroup$
    – Peter Szilas
    Dec 9 '18 at 18:30












    $begingroup$
    My opinion is that you can let it of course. There are also other answers with more advanced approach than what requested by the asker, and your one is different. Moreover some other reader could be interested to it.
    $endgroup$
    – gimusi
    Dec 9 '18 at 18:34




    $begingroup$
    My opinion is that you can let it of course. There are also other answers with more advanced approach than what requested by the asker, and your one is different. Moreover some other reader could be interested to it.
    $endgroup$
    – gimusi
    Dec 9 '18 at 18:34











    0












    $begingroup$

    With
    $$ln n<n-1$$
    then
    $$dfrac{1}{n^aln n}>dfrac{1}{n-1}n^{-a}toinfty$$
    for $-a>1$.




    Edit:
    Might be helpful:
    $$ln x=int_1^xfrac{mathrm dt}tle int_1^xfrac{t^bmathrm dt}{t}=int_1^x t^{b-1}mathrm dt=frac{x^b-1}{b}<frac{x^b}{b}$$
    for real $b>0$. Now let $b=dfrac{a}{2}$, then
    $$dfrac{1}{n^aln n}>dfrac{1}{n^frac{a}{2}n^frac{a}{2}}toinfty$$
    if $-a>0$.







    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      You don't have the case $--1le a<0$.
      $endgroup$
      – Bernard
      Dec 9 '18 at 16:01










    • $begingroup$
      the rest is proved with integrals and isn't elementary.
      $endgroup$
      – Nosrati
      Dec 9 '18 at 16:02










    • $begingroup$
      For me it's elementary inasmuch as it only uses elementary properties of integrals, as seen in high school. Or the O.P. should give us the definition of the log he/she uses.
      $endgroup$
      – Bernard
      Dec 9 '18 at 16:06










    • $begingroup$
      Sorry fro the (maybe stupid) question, but the first inequality is assumed as a given?
      $endgroup$
      – gimusi
      Dec 9 '18 at 19:19










    • $begingroup$
      As you know that one is proven with calculus, Isn't that?
      $endgroup$
      – Nosrati
      Dec 9 '18 at 19:21


















    0












    $begingroup$

    With
    $$ln n<n-1$$
    then
    $$dfrac{1}{n^aln n}>dfrac{1}{n-1}n^{-a}toinfty$$
    for $-a>1$.




    Edit:
    Might be helpful:
    $$ln x=int_1^xfrac{mathrm dt}tle int_1^xfrac{t^bmathrm dt}{t}=int_1^x t^{b-1}mathrm dt=frac{x^b-1}{b}<frac{x^b}{b}$$
    for real $b>0$. Now let $b=dfrac{a}{2}$, then
    $$dfrac{1}{n^aln n}>dfrac{1}{n^frac{a}{2}n^frac{a}{2}}toinfty$$
    if $-a>0$.







    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      You don't have the case $--1le a<0$.
      $endgroup$
      – Bernard
      Dec 9 '18 at 16:01










    • $begingroup$
      the rest is proved with integrals and isn't elementary.
      $endgroup$
      – Nosrati
      Dec 9 '18 at 16:02










    • $begingroup$
      For me it's elementary inasmuch as it only uses elementary properties of integrals, as seen in high school. Or the O.P. should give us the definition of the log he/she uses.
      $endgroup$
      – Bernard
      Dec 9 '18 at 16:06










    • $begingroup$
      Sorry fro the (maybe stupid) question, but the first inequality is assumed as a given?
      $endgroup$
      – gimusi
      Dec 9 '18 at 19:19










    • $begingroup$
      As you know that one is proven with calculus, Isn't that?
      $endgroup$
      – Nosrati
      Dec 9 '18 at 19:21
















    0












    0








    0





    $begingroup$

    With
    $$ln n<n-1$$
    then
    $$dfrac{1}{n^aln n}>dfrac{1}{n-1}n^{-a}toinfty$$
    for $-a>1$.




    Edit:
    Might be helpful:
    $$ln x=int_1^xfrac{mathrm dt}tle int_1^xfrac{t^bmathrm dt}{t}=int_1^x t^{b-1}mathrm dt=frac{x^b-1}{b}<frac{x^b}{b}$$
    for real $b>0$. Now let $b=dfrac{a}{2}$, then
    $$dfrac{1}{n^aln n}>dfrac{1}{n^frac{a}{2}n^frac{a}{2}}toinfty$$
    if $-a>0$.







    share|cite|improve this answer











    $endgroup$



    With
    $$ln n<n-1$$
    then
    $$dfrac{1}{n^aln n}>dfrac{1}{n-1}n^{-a}toinfty$$
    for $-a>1$.




    Edit:
    Might be helpful:
    $$ln x=int_1^xfrac{mathrm dt}tle int_1^xfrac{t^bmathrm dt}{t}=int_1^x t^{b-1}mathrm dt=frac{x^b-1}{b}<frac{x^b}{b}$$
    for real $b>0$. Now let $b=dfrac{a}{2}$, then
    $$dfrac{1}{n^aln n}>dfrac{1}{n^frac{a}{2}n^frac{a}{2}}toinfty$$
    if $-a>0$.








    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 9 '18 at 16:33

























    answered Dec 9 '18 at 15:47









    NosratiNosrati

    26.5k62354




    26.5k62354












    • $begingroup$
      You don't have the case $--1le a<0$.
      $endgroup$
      – Bernard
      Dec 9 '18 at 16:01










    • $begingroup$
      the rest is proved with integrals and isn't elementary.
      $endgroup$
      – Nosrati
      Dec 9 '18 at 16:02










    • $begingroup$
      For me it's elementary inasmuch as it only uses elementary properties of integrals, as seen in high school. Or the O.P. should give us the definition of the log he/she uses.
      $endgroup$
      – Bernard
      Dec 9 '18 at 16:06










    • $begingroup$
      Sorry fro the (maybe stupid) question, but the first inequality is assumed as a given?
      $endgroup$
      – gimusi
      Dec 9 '18 at 19:19










    • $begingroup$
      As you know that one is proven with calculus, Isn't that?
      $endgroup$
      – Nosrati
      Dec 9 '18 at 19:21




















    • $begingroup$
      You don't have the case $--1le a<0$.
      $endgroup$
      – Bernard
      Dec 9 '18 at 16:01










    • $begingroup$
      the rest is proved with integrals and isn't elementary.
      $endgroup$
      – Nosrati
      Dec 9 '18 at 16:02










    • $begingroup$
      For me it's elementary inasmuch as it only uses elementary properties of integrals, as seen in high school. Or the O.P. should give us the definition of the log he/she uses.
      $endgroup$
      – Bernard
      Dec 9 '18 at 16:06










    • $begingroup$
      Sorry fro the (maybe stupid) question, but the first inequality is assumed as a given?
      $endgroup$
      – gimusi
      Dec 9 '18 at 19:19










    • $begingroup$
      As you know that one is proven with calculus, Isn't that?
      $endgroup$
      – Nosrati
      Dec 9 '18 at 19:21


















    $begingroup$
    You don't have the case $--1le a<0$.
    $endgroup$
    – Bernard
    Dec 9 '18 at 16:01




    $begingroup$
    You don't have the case $--1le a<0$.
    $endgroup$
    – Bernard
    Dec 9 '18 at 16:01












    $begingroup$
    the rest is proved with integrals and isn't elementary.
    $endgroup$
    – Nosrati
    Dec 9 '18 at 16:02




    $begingroup$
    the rest is proved with integrals and isn't elementary.
    $endgroup$
    – Nosrati
    Dec 9 '18 at 16:02












    $begingroup$
    For me it's elementary inasmuch as it only uses elementary properties of integrals, as seen in high school. Or the O.P. should give us the definition of the log he/she uses.
    $endgroup$
    – Bernard
    Dec 9 '18 at 16:06




    $begingroup$
    For me it's elementary inasmuch as it only uses elementary properties of integrals, as seen in high school. Or the O.P. should give us the definition of the log he/she uses.
    $endgroup$
    – Bernard
    Dec 9 '18 at 16:06












    $begingroup$
    Sorry fro the (maybe stupid) question, but the first inequality is assumed as a given?
    $endgroup$
    – gimusi
    Dec 9 '18 at 19:19




    $begingroup$
    Sorry fro the (maybe stupid) question, but the first inequality is assumed as a given?
    $endgroup$
    – gimusi
    Dec 9 '18 at 19:19












    $begingroup$
    As you know that one is proven with calculus, Isn't that?
    $endgroup$
    – Nosrati
    Dec 9 '18 at 19:21






    $begingroup$
    As you know that one is proven with calculus, Isn't that?
    $endgroup$
    – Nosrati
    Dec 9 '18 at 19:21













    -2












    $begingroup$

    Preliminary results



    We have that $forall a>0$



    $$frac{e^{n}} {n^a} to infty$$



    indeed by ratio test



    $$frac{e^{n+1}}{(n+1)^a}frac {n^a} {e^n}to e>1$$



    then from here we have that



    $$frac{log n}{n} to 0 iff frac n {log n}to infty$$



    indeed for any $a >0$



    $$frac n {log n}ge a iff nge log n^a iff e^nge n^a$$



    Proof of the main result



    By $b=-alpha>0$ we have



    $$frac{1}{n^alpha cdot ln n }=frac{n^b}{ln n}=bfrac{n^b}{bln n}=bfrac{n^b}{ln n^b}$$



    and since for any $b>0 implies n^bto infty$ we have that



    $$bfrac{n^b}{ln n^b}to infty implies frac{1}{n^alpha cdot ln n } to infty$$






    share|cite|improve this answer











    $endgroup$


















      -2












      $begingroup$

      Preliminary results



      We have that $forall a>0$



      $$frac{e^{n}} {n^a} to infty$$



      indeed by ratio test



      $$frac{e^{n+1}}{(n+1)^a}frac {n^a} {e^n}to e>1$$



      then from here we have that



      $$frac{log n}{n} to 0 iff frac n {log n}to infty$$



      indeed for any $a >0$



      $$frac n {log n}ge a iff nge log n^a iff e^nge n^a$$



      Proof of the main result



      By $b=-alpha>0$ we have



      $$frac{1}{n^alpha cdot ln n }=frac{n^b}{ln n}=bfrac{n^b}{bln n}=bfrac{n^b}{ln n^b}$$



      and since for any $b>0 implies n^bto infty$ we have that



      $$bfrac{n^b}{ln n^b}to infty implies frac{1}{n^alpha cdot ln n } to infty$$






      share|cite|improve this answer











      $endgroup$
















        -2












        -2








        -2





        $begingroup$

        Preliminary results



        We have that $forall a>0$



        $$frac{e^{n}} {n^a} to infty$$



        indeed by ratio test



        $$frac{e^{n+1}}{(n+1)^a}frac {n^a} {e^n}to e>1$$



        then from here we have that



        $$frac{log n}{n} to 0 iff frac n {log n}to infty$$



        indeed for any $a >0$



        $$frac n {log n}ge a iff nge log n^a iff e^nge n^a$$



        Proof of the main result



        By $b=-alpha>0$ we have



        $$frac{1}{n^alpha cdot ln n }=frac{n^b}{ln n}=bfrac{n^b}{bln n}=bfrac{n^b}{ln n^b}$$



        and since for any $b>0 implies n^bto infty$ we have that



        $$bfrac{n^b}{ln n^b}to infty implies frac{1}{n^alpha cdot ln n } to infty$$






        share|cite|improve this answer











        $endgroup$



        Preliminary results



        We have that $forall a>0$



        $$frac{e^{n}} {n^a} to infty$$



        indeed by ratio test



        $$frac{e^{n+1}}{(n+1)^a}frac {n^a} {e^n}to e>1$$



        then from here we have that



        $$frac{log n}{n} to 0 iff frac n {log n}to infty$$



        indeed for any $a >0$



        $$frac n {log n}ge a iff nge log n^a iff e^nge n^a$$



        Proof of the main result



        By $b=-alpha>0$ we have



        $$frac{1}{n^alpha cdot ln n }=frac{n^b}{ln n}=bfrac{n^b}{bln n}=bfrac{n^b}{ln n^b}$$



        and since for any $b>0 implies n^bto infty$ we have that



        $$bfrac{n^b}{ln n^b}to infty implies frac{1}{n^alpha cdot ln n } to infty$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 9 '18 at 19:17

























        answered Dec 9 '18 at 15:39









        gimusigimusi

        93k84594




        93k84594






























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