Limit of $left(frac{1}{n}right)^{ln2}$












1












$begingroup$


I need prove that $left(frac{1}{n}right)^{ln2} rightarrow 0$. I think that it can be possible if I use squeeze theorem, but I have: $frac{1}{n}<left(frac{1}{n}right)^{ln2}<...$ and I don't have idea what can be a "...". Any ideas?






real-analysis







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  • 1




    $begingroup$
    maybe $frac{1}{n}^{1/2}$?
    $endgroup$
    – Yanko
    Jan 13 at 12:10










  • $begingroup$
    $n^{ln 2}to infty$, is this not clear?
    $endgroup$
    – Jakobian
    Jan 13 at 12:11










  • $begingroup$
    Why not simply use the continuity of power functions?
    $endgroup$
    – Bernard
    Jan 13 at 12:11










  • $begingroup$
    @Bernard, because it is task with a series and I can not functions in it. It is only part of my task, but it was my only problem so I did not write the whole task
    $endgroup$
    – MP3129
    Jan 13 at 12:19
















1












$begingroup$


I need prove that $left(frac{1}{n}right)^{ln2} rightarrow 0$. I think that it can be possible if I use squeeze theorem, but I have: $frac{1}{n}<left(frac{1}{n}right)^{ln2}<...$ and I don't have idea what can be a "...". Any ideas?






real-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    maybe $frac{1}{n}^{1/2}$?
    $endgroup$
    – Yanko
    Jan 13 at 12:10










  • $begingroup$
    $n^{ln 2}to infty$, is this not clear?
    $endgroup$
    – Jakobian
    Jan 13 at 12:11










  • $begingroup$
    Why not simply use the continuity of power functions?
    $endgroup$
    – Bernard
    Jan 13 at 12:11










  • $begingroup$
    @Bernard, because it is task with a series and I can not functions in it. It is only part of my task, but it was my only problem so I did not write the whole task
    $endgroup$
    – MP3129
    Jan 13 at 12:19














1












1








1





$begingroup$


I need prove that $left(frac{1}{n}right)^{ln2} rightarrow 0$. I think that it can be possible if I use squeeze theorem, but I have: $frac{1}{n}<left(frac{1}{n}right)^{ln2}<...$ and I don't have idea what can be a "...". Any ideas?






real-analysis







share|cite|improve this question











$endgroup$




I need prove that $left(frac{1}{n}right)^{ln2} rightarrow 0$. I think that it can be possible if I use squeeze theorem, but I have: $frac{1}{n}<left(frac{1}{n}right)^{ln2}<...$ and I don't have idea what can be a "...". Any ideas?






real-analysis




real-analysis






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share|cite|improve this question













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edited Jan 13 at 12:11









KM101

6,0861525




6,0861525










asked Jan 13 at 12:07









MP3129MP3129

871211




871211








  • 1




    $begingroup$
    maybe $frac{1}{n}^{1/2}$?
    $endgroup$
    – Yanko
    Jan 13 at 12:10










  • $begingroup$
    $n^{ln 2}to infty$, is this not clear?
    $endgroup$
    – Jakobian
    Jan 13 at 12:11










  • $begingroup$
    Why not simply use the continuity of power functions?
    $endgroup$
    – Bernard
    Jan 13 at 12:11










  • $begingroup$
    @Bernard, because it is task with a series and I can not functions in it. It is only part of my task, but it was my only problem so I did not write the whole task
    $endgroup$
    – MP3129
    Jan 13 at 12:19














  • 1




    $begingroup$
    maybe $frac{1}{n}^{1/2}$?
    $endgroup$
    – Yanko
    Jan 13 at 12:10










  • $begingroup$
    $n^{ln 2}to infty$, is this not clear?
    $endgroup$
    – Jakobian
    Jan 13 at 12:11










  • $begingroup$
    Why not simply use the continuity of power functions?
    $endgroup$
    – Bernard
    Jan 13 at 12:11










  • $begingroup$
    @Bernard, because it is task with a series and I can not functions in it. It is only part of my task, but it was my only problem so I did not write the whole task
    $endgroup$
    – MP3129
    Jan 13 at 12:19








1




1




$begingroup$
maybe $frac{1}{n}^{1/2}$?
$endgroup$
– Yanko
Jan 13 at 12:10




$begingroup$
maybe $frac{1}{n}^{1/2}$?
$endgroup$
– Yanko
Jan 13 at 12:10












$begingroup$
$n^{ln 2}to infty$, is this not clear?
$endgroup$
– Jakobian
Jan 13 at 12:11




$begingroup$
$n^{ln 2}to infty$, is this not clear?
$endgroup$
– Jakobian
Jan 13 at 12:11












$begingroup$
Why not simply use the continuity of power functions?
$endgroup$
– Bernard
Jan 13 at 12:11




$begingroup$
Why not simply use the continuity of power functions?
$endgroup$
– Bernard
Jan 13 at 12:11












$begingroup$
@Bernard, because it is task with a series and I can not functions in it. It is only part of my task, but it was my only problem so I did not write the whole task
$endgroup$
– MP3129
Jan 13 at 12:19




$begingroup$
@Bernard, because it is task with a series and I can not functions in it. It is only part of my task, but it was my only problem so I did not write the whole task
$endgroup$
– MP3129
Jan 13 at 12:19










2 Answers
2






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1












$begingroup$

Yes. Squeeze theorem is applicable. Note that $$e<4iff sqrt e<2$$therefore $$left({1over n}right)^{lnsqrt e}=sqrt{1over n}>left({1over n}right)^{ln 2}>0$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Observe that



    $$log 2gefrac12implies n^{log 2}ge n^{1/2}impliesfrac1{n^{log 2}}lefrac1{n^{1/2}}$$






    share|cite|improve this answer









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      2 Answers
      2






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      2 Answers
      2






      active

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      active

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      1












      $begingroup$

      Yes. Squeeze theorem is applicable. Note that $$e<4iff sqrt e<2$$therefore $$left({1over n}right)^{lnsqrt e}=sqrt{1over n}>left({1over n}right)^{ln 2}>0$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Yes. Squeeze theorem is applicable. Note that $$e<4iff sqrt e<2$$therefore $$left({1over n}right)^{lnsqrt e}=sqrt{1over n}>left({1over n}right)^{ln 2}>0$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Yes. Squeeze theorem is applicable. Note that $$e<4iff sqrt e<2$$therefore $$left({1over n}right)^{lnsqrt e}=sqrt{1over n}>left({1over n}right)^{ln 2}>0$$






          share|cite|improve this answer









          $endgroup$



          Yes. Squeeze theorem is applicable. Note that $$e<4iff sqrt e<2$$therefore $$left({1over n}right)^{lnsqrt e}=sqrt{1over n}>left({1over n}right)^{ln 2}>0$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 13 at 15:23









          Mostafa AyazMostafa Ayaz

          18.1k31040




          18.1k31040























              1












              $begingroup$

              Observe that



              $$log 2gefrac12implies n^{log 2}ge n^{1/2}impliesfrac1{n^{log 2}}lefrac1{n^{1/2}}$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Observe that



                $$log 2gefrac12implies n^{log 2}ge n^{1/2}impliesfrac1{n^{log 2}}lefrac1{n^{1/2}}$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Observe that



                  $$log 2gefrac12implies n^{log 2}ge n^{1/2}impliesfrac1{n^{log 2}}lefrac1{n^{1/2}}$$






                  share|cite|improve this answer









                  $endgroup$



                  Observe that



                  $$log 2gefrac12implies n^{log 2}ge n^{1/2}impliesfrac1{n^{log 2}}lefrac1{n^{1/2}}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 13 at 12:13









                  DonAntonioDonAntonio

                  180k1495233




                  180k1495233






























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