Can a bounded number sequence be strictly ascending?












2












$begingroup$


The title says it. Can a bounded number sequence be strictly ascending / descending?



I have a problem that tells me the sequence of fractional parts $({nx})_{ngeq 1}$ (where $x$ is given) is ascending. But I know that the sequence is bounded $[0,1)$. Thus, shouldn’t the sequence stop ascending at a point? Please show me a proof or something.






sequences-and-series







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$endgroup$








  • 1




    $begingroup$
    A series deals with summation. A sequence deals with individual elements.
    $endgroup$
    – Subhasis Biswas
    15 hours ago






  • 4




    $begingroup$
    You've received two examples of a bounded sequence that is strictly increasing. But I'm questioning your premise. For a fixed $x$, the sequence $({nx})_{nge1}$ is never strictly increasing.
    $endgroup$
    – Teepeemm
    12 hours ago
















2












$begingroup$


The title says it. Can a bounded number sequence be strictly ascending / descending?



I have a problem that tells me the sequence of fractional parts $({nx})_{ngeq 1}$ (where $x$ is given) is ascending. But I know that the sequence is bounded $[0,1)$. Thus, shouldn’t the sequence stop ascending at a point? Please show me a proof or something.






sequences-and-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    A series deals with summation. A sequence deals with individual elements.
    $endgroup$
    – Subhasis Biswas
    15 hours ago






  • 4




    $begingroup$
    You've received two examples of a bounded sequence that is strictly increasing. But I'm questioning your premise. For a fixed $x$, the sequence $({nx})_{nge1}$ is never strictly increasing.
    $endgroup$
    – Teepeemm
    12 hours ago














2












2








2





$begingroup$


The title says it. Can a bounded number sequence be strictly ascending / descending?



I have a problem that tells me the sequence of fractional parts $({nx})_{ngeq 1}$ (where $x$ is given) is ascending. But I know that the sequence is bounded $[0,1)$. Thus, shouldn’t the sequence stop ascending at a point? Please show me a proof or something.






sequences-and-series







share|cite|improve this question











$endgroup$




The title says it. Can a bounded number sequence be strictly ascending / descending?



I have a problem that tells me the sequence of fractional parts $({nx})_{ngeq 1}$ (where $x$ is given) is ascending. But I know that the sequence is bounded $[0,1)$. Thus, shouldn’t the sequence stop ascending at a point? Please show me a proof or something.






sequences-and-series




sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 14 hours ago







furfur

















asked 15 hours ago









furfurfurfur

699




699








  • 1




    $begingroup$
    A series deals with summation. A sequence deals with individual elements.
    $endgroup$
    – Subhasis Biswas
    15 hours ago






  • 4




    $begingroup$
    You've received two examples of a bounded sequence that is strictly increasing. But I'm questioning your premise. For a fixed $x$, the sequence $({nx})_{nge1}$ is never strictly increasing.
    $endgroup$
    – Teepeemm
    12 hours ago














  • 1




    $begingroup$
    A series deals with summation. A sequence deals with individual elements.
    $endgroup$
    – Subhasis Biswas
    15 hours ago






  • 4




    $begingroup$
    You've received two examples of a bounded sequence that is strictly increasing. But I'm questioning your premise. For a fixed $x$, the sequence $({nx})_{nge1}$ is never strictly increasing.
    $endgroup$
    – Teepeemm
    12 hours ago








1




1




$begingroup$
A series deals with summation. A sequence deals with individual elements.
$endgroup$
– Subhasis Biswas
15 hours ago




$begingroup$
A series deals with summation. A sequence deals with individual elements.
$endgroup$
– Subhasis Biswas
15 hours ago




4




4




$begingroup$
You've received two examples of a bounded sequence that is strictly increasing. But I'm questioning your premise. For a fixed $x$, the sequence $({nx})_{nge1}$ is never strictly increasing.
$endgroup$
– Teepeemm
12 hours ago




$begingroup$
You've received two examples of a bounded sequence that is strictly increasing. But I'm questioning your premise. For a fixed $x$, the sequence $({nx})_{nge1}$ is never strictly increasing.
$endgroup$
– Teepeemm
12 hours ago










3 Answers
3






active

oldest

votes


















21












$begingroup$


Can a bounded number sequence be strictly ascending?




Sure it can.



Hint



$0.9 ;,; 0.99 ;,; 0.999 ;,; ldots$






share|cite|improve this answer











$endgroup$





















    16












    $begingroup$

    I presume you mean sequence, not series. For example, the sequence $1 - 1/n$ is bounded and strictly increasing.






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      Yes.



      Though the mathematic series is "In mathematics, a series is, roughly speaking, a description of the operation of adding infinitely many quantities,"



      So basically the sequence of the partial sums of e.g. a geometric series
      with rate r: 0<r<1 will be an ever increasing, bounded, number. Copy pasting wikipedia:



      enter image description here



      This is also the base for Zeno's Paradoxes






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        +1 for Zeno's paradox
        $endgroup$
        – Pere
        10 hours ago











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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      21












      $begingroup$


      Can a bounded number sequence be strictly ascending?




      Sure it can.



      Hint



      $0.9 ;,; 0.99 ;,; 0.999 ;,; ldots$






      share|cite|improve this answer











      $endgroup$


















        21












        $begingroup$


        Can a bounded number sequence be strictly ascending?




        Sure it can.



        Hint



        $0.9 ;,; 0.99 ;,; 0.999 ;,; ldots$






        share|cite|improve this answer











        $endgroup$
















          21












          21








          21





          $begingroup$


          Can a bounded number sequence be strictly ascending?




          Sure it can.



          Hint



          $0.9 ;,; 0.99 ;,; 0.999 ;,; ldots$






          share|cite|improve this answer











          $endgroup$




          Can a bounded number sequence be strictly ascending?




          Sure it can.



          Hint



          $0.9 ;,; 0.99 ;,; 0.999 ;,; ldots$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 9 hours ago

























          answered 15 hours ago









          StackTDStackTD

          23.5k2154




          23.5k2154























              16












              $begingroup$

              I presume you mean sequence, not series. For example, the sequence $1 - 1/n$ is bounded and strictly increasing.






              share|cite|improve this answer









              $endgroup$


















                16












                $begingroup$

                I presume you mean sequence, not series. For example, the sequence $1 - 1/n$ is bounded and strictly increasing.






                share|cite|improve this answer









                $endgroup$
















                  16












                  16








                  16





                  $begingroup$

                  I presume you mean sequence, not series. For example, the sequence $1 - 1/n$ is bounded and strictly increasing.






                  share|cite|improve this answer









                  $endgroup$



                  I presume you mean sequence, not series. For example, the sequence $1 - 1/n$ is bounded and strictly increasing.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 15 hours ago









                  Robert IsraelRobert Israel

                  327k23216470




                  327k23216470























                      2












                      $begingroup$

                      Yes.



                      Though the mathematic series is "In mathematics, a series is, roughly speaking, a description of the operation of adding infinitely many quantities,"



                      So basically the sequence of the partial sums of e.g. a geometric series
                      with rate r: 0<r<1 will be an ever increasing, bounded, number. Copy pasting wikipedia:



                      enter image description here



                      This is also the base for Zeno's Paradoxes






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        +1 for Zeno's paradox
                        $endgroup$
                        – Pere
                        10 hours ago
















                      2












                      $begingroup$

                      Yes.



                      Though the mathematic series is "In mathematics, a series is, roughly speaking, a description of the operation of adding infinitely many quantities,"



                      So basically the sequence of the partial sums of e.g. a geometric series
                      with rate r: 0<r<1 will be an ever increasing, bounded, number. Copy pasting wikipedia:



                      enter image description here



                      This is also the base for Zeno's Paradoxes






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        +1 for Zeno's paradox
                        $endgroup$
                        – Pere
                        10 hours ago














                      2












                      2








                      2





                      $begingroup$

                      Yes.



                      Though the mathematic series is "In mathematics, a series is, roughly speaking, a description of the operation of adding infinitely many quantities,"



                      So basically the sequence of the partial sums of e.g. a geometric series
                      with rate r: 0<r<1 will be an ever increasing, bounded, number. Copy pasting wikipedia:



                      enter image description here



                      This is also the base for Zeno's Paradoxes






                      share|cite|improve this answer









                      $endgroup$



                      Yes.



                      Though the mathematic series is "In mathematics, a series is, roughly speaking, a description of the operation of adding infinitely many quantities,"



                      So basically the sequence of the partial sums of e.g. a geometric series
                      with rate r: 0<r<1 will be an ever increasing, bounded, number. Copy pasting wikipedia:



                      enter image description here



                      This is also the base for Zeno's Paradoxes







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 10 hours ago









                      ntgntg

                      1436




                      1436












                      • $begingroup$
                        +1 for Zeno's paradox
                        $endgroup$
                        – Pere
                        10 hours ago


















                      • $begingroup$
                        +1 for Zeno's paradox
                        $endgroup$
                        – Pere
                        10 hours ago
















                      $begingroup$
                      +1 for Zeno's paradox
                      $endgroup$
                      – Pere
                      10 hours ago




                      $begingroup$
                      +1 for Zeno's paradox
                      $endgroup$
                      – Pere
                      10 hours ago


















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