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Cauchy Sequence Characterized only By Directly Neighbouring Sequence Members



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Series constructed from a cauchy sequenceRelations among notions of convergenceCauchy Sequence proof with boundsProof review - (lack of rigour?) Convergent sequence iff Cauchy without Bolzano-WeierstrassProof verification regarding whether a certain property of a sequence implies that it is Cauchy.Why is the sequence $x(n) = log n$ **not** Cauchy?Mathematical Analysis Cauchy SequenceThat a sequence is Cauchy implies it's bounded.Determine if this specific sequence is a Cauchy sequenceCauchy sequence and boundedness












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$begingroup$


Let $(a_n)$ be a sequence of real numbers, for which it holds, that
$$ lim_{n rightarrow infty} lvert a_{n+1}-a_n rvert = 0. $$ Does this already imply, that $(a_n)$ is a Cauchy sequence?






limits cauchy-sequences







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let $(a_n)$ be a sequence of real numbers, for which it holds, that
    $$ lim_{n rightarrow infty} lvert a_{n+1}-a_n rvert = 0. $$ Does this already imply, that $(a_n)$ is a Cauchy sequence?






    limits cauchy-sequences







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let $(a_n)$ be a sequence of real numbers, for which it holds, that
      $$ lim_{n rightarrow infty} lvert a_{n+1}-a_n rvert = 0. $$ Does this already imply, that $(a_n)$ is a Cauchy sequence?






      limits cauchy-sequences







      share|cite|improve this question









      $endgroup$




      Let $(a_n)$ be a sequence of real numbers, for which it holds, that
      $$ lim_{n rightarrow infty} lvert a_{n+1}-a_n rvert = 0. $$ Does this already imply, that $(a_n)$ is a Cauchy sequence?






      limits cauchy-sequences




      limits cauchy-sequences






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 3 hours ago









      Joker123Joker123

      632313




      632313






















          3 Answers
          3






          active

          oldest

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          2












          $begingroup$

          Unfortunately not. Consider
          $$a_n:=sum_{i=1}^nfrac{1}{i}.$$
          We find $a_{n+1}-a_n=1/(n+1)to 0,$ but $lim_{ntoinfty}a_n=infty,$ hence ${a_n}_{ninmathbb{N}}$ is not a cauchy sequence.






          share|cite|improve this answer











          $endgroup$





















            2












            $begingroup$

            No. The sequence $a_n=sum_{k=1}^nfrac{1}{k}$ is a counterexample.






            share|cite|improve this answer









            $endgroup$





















              2












              $begingroup$

              Counterexample: $a_n = sqrt{n}$. Clearly this sequence does not converge. But
              $$
              a_{n+1} - a_{n} = sqrt{n+1} - sqrt{n} = frac{(sqrt{n+1} - sqrt{n})(sqrt{n+1} + sqrt{n})}{(sqrt{n+1} + sqrt{n})} = frac{1}{sqrt{n+1} + sqrt{n}} to 0 , .
              $$






              share|cite|improve this answer









              $endgroup$














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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                2












                $begingroup$

                Unfortunately not. Consider
                $$a_n:=sum_{i=1}^nfrac{1}{i}.$$
                We find $a_{n+1}-a_n=1/(n+1)to 0,$ but $lim_{ntoinfty}a_n=infty,$ hence ${a_n}_{ninmathbb{N}}$ is not a cauchy sequence.






                share|cite|improve this answer











                $endgroup$


















                  2












                  $begingroup$

                  Unfortunately not. Consider
                  $$a_n:=sum_{i=1}^nfrac{1}{i}.$$
                  We find $a_{n+1}-a_n=1/(n+1)to 0,$ but $lim_{ntoinfty}a_n=infty,$ hence ${a_n}_{ninmathbb{N}}$ is not a cauchy sequence.






                  share|cite|improve this answer











                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    Unfortunately not. Consider
                    $$a_n:=sum_{i=1}^nfrac{1}{i}.$$
                    We find $a_{n+1}-a_n=1/(n+1)to 0,$ but $lim_{ntoinfty}a_n=infty,$ hence ${a_n}_{ninmathbb{N}}$ is not a cauchy sequence.






                    share|cite|improve this answer











                    $endgroup$



                    Unfortunately not. Consider
                    $$a_n:=sum_{i=1}^nfrac{1}{i}.$$
                    We find $a_{n+1}-a_n=1/(n+1)to 0,$ but $lim_{ntoinfty}a_n=infty,$ hence ${a_n}_{ninmathbb{N}}$ is not a cauchy sequence.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 3 hours ago









                    HAMIDINE SOUMARE

                    2,208214




                    2,208214










                    answered 3 hours ago









                    MelodyMelody

                    1,27012




                    1,27012























                        2












                        $begingroup$

                        No. The sequence $a_n=sum_{k=1}^nfrac{1}{k}$ is a counterexample.






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          No. The sequence $a_n=sum_{k=1}^nfrac{1}{k}$ is a counterexample.






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            No. The sequence $a_n=sum_{k=1}^nfrac{1}{k}$ is a counterexample.






                            share|cite|improve this answer









                            $endgroup$



                            No. The sequence $a_n=sum_{k=1}^nfrac{1}{k}$ is a counterexample.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 3 hours ago









                            MarkMark

                            10.6k1622




                            10.6k1622























                                2












                                $begingroup$

                                Counterexample: $a_n = sqrt{n}$. Clearly this sequence does not converge. But
                                $$
                                a_{n+1} - a_{n} = sqrt{n+1} - sqrt{n} = frac{(sqrt{n+1} - sqrt{n})(sqrt{n+1} + sqrt{n})}{(sqrt{n+1} + sqrt{n})} = frac{1}{sqrt{n+1} + sqrt{n}} to 0 , .
                                $$






                                share|cite|improve this answer









                                $endgroup$


















                                  2












                                  $begingroup$

                                  Counterexample: $a_n = sqrt{n}$. Clearly this sequence does not converge. But
                                  $$
                                  a_{n+1} - a_{n} = sqrt{n+1} - sqrt{n} = frac{(sqrt{n+1} - sqrt{n})(sqrt{n+1} + sqrt{n})}{(sqrt{n+1} + sqrt{n})} = frac{1}{sqrt{n+1} + sqrt{n}} to 0 , .
                                  $$






                                  share|cite|improve this answer









                                  $endgroup$
















                                    2












                                    2








                                    2





                                    $begingroup$

                                    Counterexample: $a_n = sqrt{n}$. Clearly this sequence does not converge. But
                                    $$
                                    a_{n+1} - a_{n} = sqrt{n+1} - sqrt{n} = frac{(sqrt{n+1} - sqrt{n})(sqrt{n+1} + sqrt{n})}{(sqrt{n+1} + sqrt{n})} = frac{1}{sqrt{n+1} + sqrt{n}} to 0 , .
                                    $$






                                    share|cite|improve this answer









                                    $endgroup$



                                    Counterexample: $a_n = sqrt{n}$. Clearly this sequence does not converge. But
                                    $$
                                    a_{n+1} - a_{n} = sqrt{n+1} - sqrt{n} = frac{(sqrt{n+1} - sqrt{n})(sqrt{n+1} + sqrt{n})}{(sqrt{n+1} + sqrt{n})} = frac{1}{sqrt{n+1} + sqrt{n}} to 0 , .
                                    $$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 3 hours ago









                                    Hans EnglerHans Engler

                                    10.7k11836




                                    10.7k11836






























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