Prove that $lim_{n to infty} a_n = -infty $ if $ forall ngeq m: a_n leq a_m +2017$, and $-infty$ is a...












0












$begingroup$


Given a sequence ${a_n}_{n=1}^{infty} $ such that



$forall ngeq m: a_n leq a_m +2017$,



and $-infty$ is a substantial limit of $a_n$,



Prove that $lim_{n to infty} a_n = -infty$.



my attempt:
It's clear that: $text{lim inf}_{n to infty} (a_n) = -infty$.



I'd like to prove that $text{lim inf}_{n to infty} (a_n) = -infty$, and that will prove the claim.



there exist a sub-sequence of $a_n$,



let it be $a_{n_{k}}$ such that $$lim_{kto infty} a_{n_{k}} = -infty$$



then for every $ n geq m: $ $a_{n_{k}} leq a_{m_{k}} + 2017 $



$exists M > 0 text{ such that} forall Nin mathbb{N}: exists n geq m geq N: M leq |a_{n_{k}}| leq |a_{m_{k}} + 2017| $



I'm not sure how to continue from here in order to prove that the limit of the sub sequence is the limit of the sequence.






real-analysis calculus sequences-and-series convergence







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Given a sequence ${a_n}_{n=1}^{infty} $ such that



    $forall ngeq m: a_n leq a_m +2017$,



    and $-infty$ is a substantial limit of $a_n$,



    Prove that $lim_{n to infty} a_n = -infty$.



    my attempt:
    It's clear that: $text{lim inf}_{n to infty} (a_n) = -infty$.



    I'd like to prove that $text{lim inf}_{n to infty} (a_n) = -infty$, and that will prove the claim.



    there exist a sub-sequence of $a_n$,



    let it be $a_{n_{k}}$ such that $$lim_{kto infty} a_{n_{k}} = -infty$$



    then for every $ n geq m: $ $a_{n_{k}} leq a_{m_{k}} + 2017 $



    $exists M > 0 text{ such that} forall Nin mathbb{N}: exists n geq m geq N: M leq |a_{n_{k}}| leq |a_{m_{k}} + 2017| $



    I'm not sure how to continue from here in order to prove that the limit of the sub sequence is the limit of the sequence.






    real-analysis calculus sequences-and-series convergence







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Given a sequence ${a_n}_{n=1}^{infty} $ such that



      $forall ngeq m: a_n leq a_m +2017$,



      and $-infty$ is a substantial limit of $a_n$,



      Prove that $lim_{n to infty} a_n = -infty$.



      my attempt:
      It's clear that: $text{lim inf}_{n to infty} (a_n) = -infty$.



      I'd like to prove that $text{lim inf}_{n to infty} (a_n) = -infty$, and that will prove the claim.



      there exist a sub-sequence of $a_n$,



      let it be $a_{n_{k}}$ such that $$lim_{kto infty} a_{n_{k}} = -infty$$



      then for every $ n geq m: $ $a_{n_{k}} leq a_{m_{k}} + 2017 $



      $exists M > 0 text{ such that} forall Nin mathbb{N}: exists n geq m geq N: M leq |a_{n_{k}}| leq |a_{m_{k}} + 2017| $



      I'm not sure how to continue from here in order to prove that the limit of the sub sequence is the limit of the sequence.






      real-analysis calculus sequences-and-series convergence







      share|cite|improve this question











      $endgroup$




      Given a sequence ${a_n}_{n=1}^{infty} $ such that



      $forall ngeq m: a_n leq a_m +2017$,



      and $-infty$ is a substantial limit of $a_n$,



      Prove that $lim_{n to infty} a_n = -infty$.



      my attempt:
      It's clear that: $text{lim inf}_{n to infty} (a_n) = -infty$.



      I'd like to prove that $text{lim inf}_{n to infty} (a_n) = -infty$, and that will prove the claim.



      there exist a sub-sequence of $a_n$,



      let it be $a_{n_{k}}$ such that $$lim_{kto infty} a_{n_{k}} = -infty$$



      then for every $ n geq m: $ $a_{n_{k}} leq a_{m_{k}} + 2017 $



      $exists M > 0 text{ such that} forall Nin mathbb{N}: exists n geq m geq N: M leq |a_{n_{k}}| leq |a_{m_{k}} + 2017| $



      I'm not sure how to continue from here in order to prove that the limit of the sub sequence is the limit of the sequence.






      real-analysis calculus sequences-and-series convergence




      real-analysis calculus sequences-and-series convergence






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 5 '18 at 17:14









      Martin Sleziak

      44.8k9118272




      44.8k9118272










      asked Dec 5 '18 at 16:12









      JnevenJneven

      826322




      826322






















          2 Answers
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          $begingroup$

          Fix $m$ then $$limsup_{ntoinfty } a_n leq limsup_{ntoinfty } (a_m +2017 )= a_m +2017$$
          Now $$limsup_{ntoinfty } a_n=liminf_{mtoinfty } ( limsup_{ntoinfty } a_n) leq liminf_{mtoinfty } ( a_m +2017)=2017+liminf_{mtoinfty } a_m =-infty$$






          share|cite|improve this answer









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            1












            $begingroup$

            If $liminf_{ktoinfty} a_{n_k} = -infty$ then for every $M$ there is some $K$ such that $a_{n_k} < M$ for every $k geq K$, and by the condition stated, $a_n leq a_{n_k} + 2017 < M + 2017$ for every $ngeq n_k$. If we pick $M = -N-2017$ then $a_n < -N$ for every $ngeq n_k$, so the tail of the sequence can be made arbitrarily small for $n$ large enough, which is the definition of $lim_{ntoinfty} a_n = -infty$.






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              2 Answers
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              2 Answers
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              $begingroup$

              Fix $m$ then $$limsup_{ntoinfty } a_n leq limsup_{ntoinfty } (a_m +2017 )= a_m +2017$$
              Now $$limsup_{ntoinfty } a_n=liminf_{mtoinfty } ( limsup_{ntoinfty } a_n) leq liminf_{mtoinfty } ( a_m +2017)=2017+liminf_{mtoinfty } a_m =-infty$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Fix $m$ then $$limsup_{ntoinfty } a_n leq limsup_{ntoinfty } (a_m +2017 )= a_m +2017$$
                Now $$limsup_{ntoinfty } a_n=liminf_{mtoinfty } ( limsup_{ntoinfty } a_n) leq liminf_{mtoinfty } ( a_m +2017)=2017+liminf_{mtoinfty } a_m =-infty$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Fix $m$ then $$limsup_{ntoinfty } a_n leq limsup_{ntoinfty } (a_m +2017 )= a_m +2017$$
                  Now $$limsup_{ntoinfty } a_n=liminf_{mtoinfty } ( limsup_{ntoinfty } a_n) leq liminf_{mtoinfty } ( a_m +2017)=2017+liminf_{mtoinfty } a_m =-infty$$






                  share|cite|improve this answer









                  $endgroup$



                  Fix $m$ then $$limsup_{ntoinfty } a_n leq limsup_{ntoinfty } (a_m +2017 )= a_m +2017$$
                  Now $$limsup_{ntoinfty } a_n=liminf_{mtoinfty } ( limsup_{ntoinfty } a_n) leq liminf_{mtoinfty } ( a_m +2017)=2017+liminf_{mtoinfty } a_m =-infty$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 5 '18 at 16:52









                  MotylaNogaTomkaMazuraMotylaNogaTomkaMazura

                  6,562917




                  6,562917























                      1












                      $begingroup$

                      If $liminf_{ktoinfty} a_{n_k} = -infty$ then for every $M$ there is some $K$ such that $a_{n_k} < M$ for every $k geq K$, and by the condition stated, $a_n leq a_{n_k} + 2017 < M + 2017$ for every $ngeq n_k$. If we pick $M = -N-2017$ then $a_n < -N$ for every $ngeq n_k$, so the tail of the sequence can be made arbitrarily small for $n$ large enough, which is the definition of $lim_{ntoinfty} a_n = -infty$.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        If $liminf_{ktoinfty} a_{n_k} = -infty$ then for every $M$ there is some $K$ such that $a_{n_k} < M$ for every $k geq K$, and by the condition stated, $a_n leq a_{n_k} + 2017 < M + 2017$ for every $ngeq n_k$. If we pick $M = -N-2017$ then $a_n < -N$ for every $ngeq n_k$, so the tail of the sequence can be made arbitrarily small for $n$ large enough, which is the definition of $lim_{ntoinfty} a_n = -infty$.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          If $liminf_{ktoinfty} a_{n_k} = -infty$ then for every $M$ there is some $K$ such that $a_{n_k} < M$ for every $k geq K$, and by the condition stated, $a_n leq a_{n_k} + 2017 < M + 2017$ for every $ngeq n_k$. If we pick $M = -N-2017$ then $a_n < -N$ for every $ngeq n_k$, so the tail of the sequence can be made arbitrarily small for $n$ large enough, which is the definition of $lim_{ntoinfty} a_n = -infty$.






                          share|cite|improve this answer









                          $endgroup$



                          If $liminf_{ktoinfty} a_{n_k} = -infty$ then for every $M$ there is some $K$ such that $a_{n_k} < M$ for every $k geq K$, and by the condition stated, $a_n leq a_{n_k} + 2017 < M + 2017$ for every $ngeq n_k$. If we pick $M = -N-2017$ then $a_n < -N$ for every $ngeq n_k$, so the tail of the sequence can be made arbitrarily small for $n$ large enough, which is the definition of $lim_{ntoinfty} a_n = -infty$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 5 '18 at 16:48









                          mlerma54mlerma54

                          1,177148




                          1,177148






























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