Cauchy Sequence Characterized only By Directly Neighbouring Sequence Members












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









$endgroup$

















    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?










    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?










      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






      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

          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












            $begingroup$

            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









              $endgroup$














                Your Answer








                StackExchange.ready(function() {
                var channelOptions = {
                tags: "".split(" "),
                id: "69"
                };
                initTagRenderer("".split(" "), "".split(" "), channelOptions);

                StackExchange.using("externalEditor", function() {
                // Have to fire editor after snippets, if snippets enabled
                if (StackExchange.settings.snippets.snippetsEnabled) {
                StackExchange.using("snippets", function() {
                createEditor();
                });
                }
                else {
                createEditor();
                }
                });

                function createEditor() {
                StackExchange.prepareEditor({
                heartbeatType: 'answer',
                autoActivateHeartbeat: false,
                convertImagesToLinks: true,
                noModals: true,
                showLowRepImageUploadWarning: true,
                reputationToPostImages: 10,
                bindNavPrevention: true,
                postfix: "",
                imageUploader: {
                brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
                contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
                allowUrls: true
                },
                noCode: true, onDemand: true,
                discardSelector: ".discard-answer"
                ,immediatelyShowMarkdownHelp:true
                });


                }
                });














                draft saved

                draft discarded


















                StackExchange.ready(
                function () {
                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3188087%2fcauchy-sequence-characterized-only-by-directly-neighbouring-sequence-members%23new-answer', 'question_page');
                }
                );

                Post as a guest















                Required, but never shown

























                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 2 hours ago









                                    Hans EnglerHans Engler

                                    10.7k11836




                                    10.7k11836






























                                        draft saved

                                        draft discarded




















































                                        Thanks for contributing an answer to Mathematics Stack Exchange!


                                        • Please be sure to answer the question. Provide details and share your research!

                                        But avoid



                                        • Asking for help, clarification, or responding to other answers.

                                        • Making statements based on opinion; back them up with references or personal experience.


                                        Use MathJax to format equations. MathJax reference.


                                        To learn more, see our tips on writing great answers.




                                        draft saved


                                        draft discarded














                                        StackExchange.ready(
                                        function () {
                                        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3188087%2fcauchy-sequence-characterized-only-by-directly-neighbouring-sequence-members%23new-answer', 'question_page');
                                        }
                                        );

                                        Post as a guest















                                        Required, but never shown





















































                                        Required, but never shown














                                        Required, but never shown












                                        Required, but never shown







                                        Required, but never shown

































                                        Required, but never shown














                                        Required, but never shown












                                        Required, but never shown







                                        Required, but never shown







                                        Popular posts from this blog

                                        Liste der Baudenkmale in Friedland (Mecklenburg)

                                        Single-Malt-Whisky

                                        Czorneboh