Determine the set of the limit points of the sequence $a_n$ ,where $a_n$ = $frac2n^27$ - $[frac2n^2 7]$

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Determine the set of the limit points of the sequence $a_n$ ,where $a_n$ = $frac2n^27$ - $[frac2n^2 7]$



My answer :i thinks $0$ will be the limit points because $a_n$ = $frac2n^27$ - $[frac2n^2 7] ge 0$



is it true ?



Any Hints /solution will be appreciated



thanks i advance....




real-analysis




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  • 6




    If by $;[ ];$ you mean the floor function then the question is weird, as $;[2n^2]=2n^2,,,,[7]=7;$ ....What did you really mean?
    – DonAntonio
    Jul 22 at 16:52










  • @DonAntonio..see i have edited...that was my Typo mistakes
    – Messi fifa
    Jul 22 at 18:08














up vote
0
down vote

favorite












Determine the set of the limit points of the sequence $a_n$ ,where $a_n$ = $frac2n^27$ - $[frac2n^2 7]$



My answer :i thinks $0$ will be the limit points because $a_n$ = $frac2n^27$ - $[frac2n^2 7] ge 0$



is it true ?



Any Hints /solution will be appreciated



thanks i advance....




real-analysis




share|cite|improve this question

















  • 6




    If by $;[ ];$ you mean the floor function then the question is weird, as $;[2n^2]=2n^2,,,,[7]=7;$ ....What did you really mean?
    – DonAntonio
    Jul 22 at 16:52










  • @DonAntonio..see i have edited...that was my Typo mistakes
    – Messi fifa
    Jul 22 at 18:08












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Determine the set of the limit points of the sequence $a_n$ ,where $a_n$ = $frac2n^27$ - $[frac2n^2 7]$



My answer :i thinks $0$ will be the limit points because $a_n$ = $frac2n^27$ - $[frac2n^2 7] ge 0$



is it true ?



Any Hints /solution will be appreciated



thanks i advance....




real-analysis




share|cite|improve this question













Determine the set of the limit points of the sequence $a_n$ ,where $a_n$ = $frac2n^27$ - $[frac2n^2 7]$



My answer :i thinks $0$ will be the limit points because $a_n$ = $frac2n^27$ - $[frac2n^2 7] ge 0$



is it true ?



Any Hints /solution will be appreciated



thanks i advance....





real-analysis





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 22 at 18:06
























asked Jul 22 at 16:49









Messi fifa

1718




1718







  • 6




    If by $;[ ];$ you mean the floor function then the question is weird, as $;[2n^2]=2n^2,,,,[7]=7;$ ....What did you really mean?
    – DonAntonio
    Jul 22 at 16:52










  • @DonAntonio..see i have edited...that was my Typo mistakes
    – Messi fifa
    Jul 22 at 18:08












  • 6




    If by $;[ ];$ you mean the floor function then the question is weird, as $;[2n^2]=2n^2,,,,[7]=7;$ ....What did you really mean?
    – DonAntonio
    Jul 22 at 16:52










  • @DonAntonio..see i have edited...that was my Typo mistakes
    – Messi fifa
    Jul 22 at 18:08







6




6




If by $;[ ];$ you mean the floor function then the question is weird, as $;[2n^2]=2n^2,,,,[7]=7;$ ....What did you really mean?
– DonAntonio
Jul 22 at 16:52




If by $;[ ];$ you mean the floor function then the question is weird, as $;[2n^2]=2n^2,,,,[7]=7;$ ....What did you really mean?
– DonAntonio
Jul 22 at 16:52












@DonAntonio..see i have edited...that was my Typo mistakes
– Messi fifa
Jul 22 at 18:08




@DonAntonio..see i have edited...that was my Typo mistakes
– Messi fifa
Jul 22 at 18:08










1 Answer
1






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1
down vote



accepted










Notation: $q-[q]=q$.



Obs 1: if $n=7k+r$, with $rin0,1,2,3,4,5,6$, then $leftfracn7right=r/7$.



Obs 2: if $n=7k+r$, with $rin0,1,2,3,4,5,6$, then $2n^2equiv 0,2,1,4,4,1,2mod 7$, respectively.



With this, we can note that $a_n=leftfrac2n^27right$ can be only $0,1/7,2/7, 4/7$ and this values can be reached endless times interspersed.



Then, their are the limit points



Edited for more clarification: This subsequences are constant



  • $a_7k=0$ for all $kinBbbN$.

  • $a_7k+2=1/7$ for all $kinBbbN$.

  • $a_7k+1=2/7$ for all $kinBbbN$.

  • $a_7k+3=4/7$ for all $kinBbbN$.





share|cite|improve this answer























  • thanks u Martin
    – Messi fifa
    Jul 22 at 18:29










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










Notation: $q-[q]=q$.



Obs 1: if $n=7k+r$, with $rin0,1,2,3,4,5,6$, then $leftfracn7right=r/7$.



Obs 2: if $n=7k+r$, with $rin0,1,2,3,4,5,6$, then $2n^2equiv 0,2,1,4,4,1,2mod 7$, respectively.



With this, we can note that $a_n=leftfrac2n^27right$ can be only $0,1/7,2/7, 4/7$ and this values can be reached endless times interspersed.



Then, their are the limit points



Edited for more clarification: This subsequences are constant



  • $a_7k=0$ for all $kinBbbN$.

  • $a_7k+2=1/7$ for all $kinBbbN$.

  • $a_7k+1=2/7$ for all $kinBbbN$.

  • $a_7k+3=4/7$ for all $kinBbbN$.





share|cite|improve this answer























  • thanks u Martin
    – Messi fifa
    Jul 22 at 18:29














up vote
1
down vote



accepted










Notation: $q-[q]=q$.



Obs 1: if $n=7k+r$, with $rin0,1,2,3,4,5,6$, then $leftfracn7right=r/7$.



Obs 2: if $n=7k+r$, with $rin0,1,2,3,4,5,6$, then $2n^2equiv 0,2,1,4,4,1,2mod 7$, respectively.



With this, we can note that $a_n=leftfrac2n^27right$ can be only $0,1/7,2/7, 4/7$ and this values can be reached endless times interspersed.



Then, their are the limit points



Edited for more clarification: This subsequences are constant



  • $a_7k=0$ for all $kinBbbN$.

  • $a_7k+2=1/7$ for all $kinBbbN$.

  • $a_7k+1=2/7$ for all $kinBbbN$.

  • $a_7k+3=4/7$ for all $kinBbbN$.





share|cite|improve this answer























  • thanks u Martin
    – Messi fifa
    Jul 22 at 18:29












up vote
1
down vote



accepted







up vote
1
down vote



accepted






Notation: $q-[q]=q$.



Obs 1: if $n=7k+r$, with $rin0,1,2,3,4,5,6$, then $leftfracn7right=r/7$.



Obs 2: if $n=7k+r$, with $rin0,1,2,3,4,5,6$, then $2n^2equiv 0,2,1,4,4,1,2mod 7$, respectively.



With this, we can note that $a_n=leftfrac2n^27right$ can be only $0,1/7,2/7, 4/7$ and this values can be reached endless times interspersed.



Then, their are the limit points



Edited for more clarification: This subsequences are constant



  • $a_7k=0$ for all $kinBbbN$.

  • $a_7k+2=1/7$ for all $kinBbbN$.

  • $a_7k+1=2/7$ for all $kinBbbN$.

  • $a_7k+3=4/7$ for all $kinBbbN$.





share|cite|improve this answer















Notation: $q-[q]=q$.



Obs 1: if $n=7k+r$, with $rin0,1,2,3,4,5,6$, then $leftfracn7right=r/7$.



Obs 2: if $n=7k+r$, with $rin0,1,2,3,4,5,6$, then $2n^2equiv 0,2,1,4,4,1,2mod 7$, respectively.



With this, we can note that $a_n=leftfrac2n^27right$ can be only $0,1/7,2/7, 4/7$ and this values can be reached endless times interspersed.



Then, their are the limit points



Edited for more clarification: This subsequences are constant



  • $a_7k=0$ for all $kinBbbN$.

  • $a_7k+2=1/7$ for all $kinBbbN$.

  • $a_7k+1=2/7$ for all $kinBbbN$.

  • $a_7k+3=4/7$ for all $kinBbbN$.






share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 22 at 18:30


























answered Jul 22 at 18:25









Martín Vacas Vignolo

3,418421




3,418421











  • thanks u Martin
    – Messi fifa
    Jul 22 at 18:29
















  • thanks u Martin
    – Messi fifa
    Jul 22 at 18:29















thanks u Martin
– Messi fifa
Jul 22 at 18:29




thanks u Martin
– Messi fifa
Jul 22 at 18:29












 

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