Cauchy Sequence Characterized only By Directly Neighbouring Sequence Members












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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?










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    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?










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      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?










      share|cite|improve this question









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      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






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      asked 3 hours ago









      Joker123Joker123

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          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.






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






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






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                3 Answers
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                3 Answers
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                active

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                active

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                $begingroup$

                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.






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                  $begingroup$

                  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











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                    2












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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











                    $endgroup$



                    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.







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                    edited 3 hours ago









                    HAMIDINE SOUMARE

                    2,208214




                    2,208214










                    answered 3 hours ago









                    MelodyMelody

                    1,27012




                    1,27012























                        2












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






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                          2












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






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                            2












                            2








                            2





                            $begingroup$

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






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







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                            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
                                $$
                                a_{n+1} - a_{n} = sqrt{n+1} - sqrt{n} = frac{(sqrt{n+1} - sqrt{n})(sqrt{n+1} + sqrt{n})}{(sqrt{n+1} + sqrt{n})} = frac{1}{sqrt{n+1} + sqrt{n}} to 0 , .
                                $$






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                                  2












                                  $begingroup$

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






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                                    2












                                    2








                                    2





                                    $begingroup$

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






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







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                                    answered 2 hours ago









                                    Hans EnglerHans Engler

                                    10.7k11836




                                    10.7k11836






























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