Limit Inferior of Sequence $x_n = 2^n$

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I am doing a problem which said that the limit inferior of sequence $x_n = 2^n$ does not exist. But isn't that the limit inferior of this sequence is $2$, or there are different definitions of limit inferior? Thanks!





limits




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  • The $liminf$ is the smallest value that some subsequence converges to. Here there is no convergent subsequence, so there is no $liminf$ (if you allow for $+infty$ then the $liminf$ is $+infty$ since every subsequence goes to $+infty$).
    – User8128
    Aug 18 at 3:38














up vote
0
down vote

favorite













I am doing a problem which said that the limit inferior of sequence $x_n = 2^n$ does not exist. But isn't that the limit inferior of this sequence is $2$, or there are different definitions of limit inferior? Thanks!





limits




share|cite|improve this question






















  • The $liminf$ is the smallest value that some subsequence converges to. Here there is no convergent subsequence, so there is no $liminf$ (if you allow for $+infty$ then the $liminf$ is $+infty$ since every subsequence goes to $+infty$).
    – User8128
    Aug 18 at 3:38












up vote
0
down vote

favorite









up vote
0
down vote

favorite












I am doing a problem which said that the limit inferior of sequence $x_n = 2^n$ does not exist. But isn't that the limit inferior of this sequence is $2$, or there are different definitions of limit inferior? Thanks!





limits




share|cite|improve this question















I am doing a problem which said that the limit inferior of sequence $x_n = 2^n$ does not exist. But isn't that the limit inferior of this sequence is $2$, or there are different definitions of limit inferior? Thanks!






limits





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 18 at 4:02









Math Lover

12.6k21232




12.6k21232










asked Aug 18 at 3:29









Cathy

1207




1207











  • The $liminf$ is the smallest value that some subsequence converges to. Here there is no convergent subsequence, so there is no $liminf$ (if you allow for $+infty$ then the $liminf$ is $+infty$ since every subsequence goes to $+infty$).
    – User8128
    Aug 18 at 3:38
















  • The $liminf$ is the smallest value that some subsequence converges to. Here there is no convergent subsequence, so there is no $liminf$ (if you allow for $+infty$ then the $liminf$ is $+infty$ since every subsequence goes to $+infty$).
    – User8128
    Aug 18 at 3:38















The $liminf$ is the smallest value that some subsequence converges to. Here there is no convergent subsequence, so there is no $liminf$ (if you allow for $+infty$ then the $liminf$ is $+infty$ since every subsequence goes to $+infty$).
– User8128
Aug 18 at 3:38




The $liminf$ is the smallest value that some subsequence converges to. Here there is no convergent subsequence, so there is no $liminf$ (if you allow for $+infty$ then the $liminf$ is $+infty$ since every subsequence goes to $+infty$).
– User8128
Aug 18 at 3:38










2 Answers
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The limit inferior is defined as follows



$$lim_n to infty inf x_n := lim_n to infty big(inf_mge n hspace4pt x_mbig)$$



here the limit inferior is divergent thus does not exit.






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    $beta_m=infx_n:nge m$. $beta_m=2^1,2^2,....$ We can easily prove that $beta_m$ is divergent. Hence, Limit inferior doesn't exist.






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      2 Answers
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      active

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

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      up vote
      3
      down vote



      accepted










      The limit inferior is defined as follows



      $$lim_n to infty inf x_n := lim_n to infty big(inf_mge n hspace4pt x_mbig)$$



      here the limit inferior is divergent thus does not exit.






      share|cite|improve this answer


























        up vote
        3
        down vote



        accepted










        The limit inferior is defined as follows



        $$lim_n to infty inf x_n := lim_n to infty big(inf_mge n hspace4pt x_mbig)$$



        here the limit inferior is divergent thus does not exit.






        share|cite|improve this answer
























          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          The limit inferior is defined as follows



          $$lim_n to infty inf x_n := lim_n to infty big(inf_mge n hspace4pt x_mbig)$$



          here the limit inferior is divergent thus does not exit.






          share|cite|improve this answer














          The limit inferior is defined as follows



          $$lim_n to infty inf x_n := lim_n to infty big(inf_mge n hspace4pt x_mbig)$$



          here the limit inferior is divergent thus does not exit.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 18 at 4:01









          Math Lover

          12.6k21232




          12.6k21232










          answered Aug 18 at 3:33









          James

          2,188619




          2,188619




















              up vote
              1
              down vote













              $beta_m=infx_n:nge m$. $beta_m=2^1,2^2,....$ We can easily prove that $beta_m$ is divergent. Hence, Limit inferior doesn't exist.






              share|cite|improve this answer
























                up vote
                1
                down vote













                $beta_m=infx_n:nge m$. $beta_m=2^1,2^2,....$ We can easily prove that $beta_m$ is divergent. Hence, Limit inferior doesn't exist.






                share|cite|improve this answer






















                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  $beta_m=infx_n:nge m$. $beta_m=2^1,2^2,....$ We can easily prove that $beta_m$ is divergent. Hence, Limit inferior doesn't exist.






                  share|cite|improve this answer












                  $beta_m=infx_n:nge m$. $beta_m=2^1,2^2,....$ We can easily prove that $beta_m$ is divergent. Hence, Limit inferior doesn't exist.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 18 at 3:43









                  N. Maneesh

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