Verification of proof of nonconvergenceProof Verification: Cauchy Sequences are convergent.I don't understand...

The need of reserving one's ability in job interviews

How to roleplay my character's ethics according to the DM when I don't understand those ethics?

Are Wave equations equivalent to Maxwell equations in free space?

Is there a math equivalent to the conditional ternary operator?

Remove object from array based on array of some property of that object

Searching for a string that contains the file name

Learning to quickly identify valid fingering for piano?

Convert an array of objects to array of the objects' values

How does insurance birth control work in the United States?

Did Amazon pay $0 in taxes last year?

Is being socially reclusive okay for a graduate student?

What is better: yes / no radio, or simple checkbox?

Adding thousand separator to various number types

How can friction do no work in case of pure rolling?

Has a sovereign Communist government ever run, and conceded loss, on a fair election?

What's the best tool for cutting holes into duct work?

Dukha vs legitimate need

Why do phishing e-mails use faked e-mail addresses instead of the real one?

In the world of The Matrix, what is "popping"?

Number of folds to form a cube, using a square paper?

Meshing the cow

How do you make a gun that shoots melee weapons and/or swords?

What is the oldest European royal house?

How will Occam's Razor principle work in Machine learning



Verification of proof of nonconvergence


Proof Verification: Cauchy Sequences are convergent.I don't understand the proof that $Bbb{R}$ is complete with respect to Cauchy sequences.Proof verification regarding whether a certain property of a sequence implies that it is Cauchy.Proof verification: convergent series with a finite number of negative terms is Absolutely ConvergentProof Verification: Regarding equivalent sequences and Cauchy sequences, Tao Analysis I: Exercise 5.2.1.A convergent sequence is bounded: Proof VerificationProof verification - Cauchy completeness of $mathbb{R}$Cauchy sequence exercise question (Exercise 2.6.4 Abbott analysis)Let $(a_n)$ and $(b_n)$ be Cauchy sequences of rationals. Then $(a_nb_n)$ is Cauchy sequenceLet there be a sequence such that the distance between two consecutive terms converges to 0. Must this sequence converge?













0












$begingroup$


I'm not sure if my proof is sound or not, I was wondering if anyone could verify.



Prove that the sequence ${a_n}_n $ given by $a_n = frac{(-1)^n;n+1}{3n}$ does not converge.



Proof: Let $epsilon = frac 23$ and $N in Bbb R$. Let $k in Bbb N$ such that $k gt N$ and suppose $m = k +1$ and $n = k$ such that $n $ is even. Then, $m,n gt N$ and $|a_m-a_n| = |frac {(-1)^n;n+1}{3n}-frac{(-1)^{n+1};n+2}{3(n+1)}|.$ Since $n$ is even, $m$ must be odd and so $|a_m-a_n|=|frac{n+1}{3n}+frac{n+2}{3(n+1)}| = |frac{(n+1)^2+n(n+2)}{3n(n+1)}|=|frac{2n^2+4n+1}{3n(n+1)}|gtfrac{2n^2+4n}{3n(n+1)}=frac{2n(n+2)}{3n(n+1)}gtfrac{2n(n+1)}{3n(n+1)}=frac23 = epsilon.$ Hence, $|a_m-a_n| ge epsilon $ for all $m,n gt N$ and so ${a_n}_n$ is not Cauchy and does not converge. $Box$



Any help would be appreciated, thanks!






real-analysis proof-verification convergence cauchy-sequences







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I'm not sure if my proof is sound or not, I was wondering if anyone could verify.



    Prove that the sequence ${a_n}_n $ given by $a_n = frac{(-1)^n;n+1}{3n}$ does not converge.



    Proof: Let $epsilon = frac 23$ and $N in Bbb R$. Let $k in Bbb N$ such that $k gt N$ and suppose $m = k +1$ and $n = k$ such that $n $ is even. Then, $m,n gt N$ and $|a_m-a_n| = |frac {(-1)^n;n+1}{3n}-frac{(-1)^{n+1};n+2}{3(n+1)}|.$ Since $n$ is even, $m$ must be odd and so $|a_m-a_n|=|frac{n+1}{3n}+frac{n+2}{3(n+1)}| = |frac{(n+1)^2+n(n+2)}{3n(n+1)}|=|frac{2n^2+4n+1}{3n(n+1)}|gtfrac{2n^2+4n}{3n(n+1)}=frac{2n(n+2)}{3n(n+1)}gtfrac{2n(n+1)}{3n(n+1)}=frac23 = epsilon.$ Hence, $|a_m-a_n| ge epsilon $ for all $m,n gt N$ and so ${a_n}_n$ is not Cauchy and does not converge. $Box$



    Any help would be appreciated, thanks!






    real-analysis proof-verification convergence cauchy-sequences







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I'm not sure if my proof is sound or not, I was wondering if anyone could verify.



      Prove that the sequence ${a_n}_n $ given by $a_n = frac{(-1)^n;n+1}{3n}$ does not converge.



      Proof: Let $epsilon = frac 23$ and $N in Bbb R$. Let $k in Bbb N$ such that $k gt N$ and suppose $m = k +1$ and $n = k$ such that $n $ is even. Then, $m,n gt N$ and $|a_m-a_n| = |frac {(-1)^n;n+1}{3n}-frac{(-1)^{n+1};n+2}{3(n+1)}|.$ Since $n$ is even, $m$ must be odd and so $|a_m-a_n|=|frac{n+1}{3n}+frac{n+2}{3(n+1)}| = |frac{(n+1)^2+n(n+2)}{3n(n+1)}|=|frac{2n^2+4n+1}{3n(n+1)}|gtfrac{2n^2+4n}{3n(n+1)}=frac{2n(n+2)}{3n(n+1)}gtfrac{2n(n+1)}{3n(n+1)}=frac23 = epsilon.$ Hence, $|a_m-a_n| ge epsilon $ for all $m,n gt N$ and so ${a_n}_n$ is not Cauchy and does not converge. $Box$



      Any help would be appreciated, thanks!






      real-analysis proof-verification convergence cauchy-sequences







      share|cite|improve this question











      $endgroup$




      I'm not sure if my proof is sound or not, I was wondering if anyone could verify.



      Prove that the sequence ${a_n}_n $ given by $a_n = frac{(-1)^n;n+1}{3n}$ does not converge.



      Proof: Let $epsilon = frac 23$ and $N in Bbb R$. Let $k in Bbb N$ such that $k gt N$ and suppose $m = k +1$ and $n = k$ such that $n $ is even. Then, $m,n gt N$ and $|a_m-a_n| = |frac {(-1)^n;n+1}{3n}-frac{(-1)^{n+1};n+2}{3(n+1)}|.$ Since $n$ is even, $m$ must be odd and so $|a_m-a_n|=|frac{n+1}{3n}+frac{n+2}{3(n+1)}| = |frac{(n+1)^2+n(n+2)}{3n(n+1)}|=|frac{2n^2+4n+1}{3n(n+1)}|gtfrac{2n^2+4n}{3n(n+1)}=frac{2n(n+2)}{3n(n+1)}gtfrac{2n(n+1)}{3n(n+1)}=frac23 = epsilon.$ Hence, $|a_m-a_n| ge epsilon $ for all $m,n gt N$ and so ${a_n}_n$ is not Cauchy and does not converge. $Box$



      Any help would be appreciated, thanks!






      real-analysis proof-verification convergence cauchy-sequences




      real-analysis proof-verification convergence cauchy-sequences






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 6 hours ago









      José Carlos Santos

      165k22132235




      165k22132235










      asked 6 hours ago









      B RetnikB Retnik

      534




      534






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          That proof is correct. I think that there is no need to introduce the letter $k$; you could just say that $m=n+1$ (in fact, that's what you actually do).



          You could also prove that that sequence has a subsequence which converges to $frac13$ and another one which converges to $-frac13$. That would also prove that the sequence is not convergent.






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            You have a much simpler way to write that. Indeed, suppose that $a_n rightarrow l$. Then
            $$frac{3n}{n+1}a_n rightarrow 3l$$



            But $frac{3n}{n+1}a_n = (-1)^n$, therefore it does not converge.






            share|cite|improve this answer









            $endgroup$













              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              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%2f3138718%2fverification-of-proof-of-nonconvergence%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              That proof is correct. I think that there is no need to introduce the letter $k$; you could just say that $m=n+1$ (in fact, that's what you actually do).



              You could also prove that that sequence has a subsequence which converges to $frac13$ and another one which converges to $-frac13$. That would also prove that the sequence is not convergent.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                That proof is correct. I think that there is no need to introduce the letter $k$; you could just say that $m=n+1$ (in fact, that's what you actually do).



                You could also prove that that sequence has a subsequence which converges to $frac13$ and another one which converges to $-frac13$. That would also prove that the sequence is not convergent.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  That proof is correct. I think that there is no need to introduce the letter $k$; you could just say that $m=n+1$ (in fact, that's what you actually do).



                  You could also prove that that sequence has a subsequence which converges to $frac13$ and another one which converges to $-frac13$. That would also prove that the sequence is not convergent.






                  share|cite|improve this answer











                  $endgroup$



                  That proof is correct. I think that there is no need to introduce the letter $k$; you could just say that $m=n+1$ (in fact, that's what you actually do).



                  You could also prove that that sequence has a subsequence which converges to $frac13$ and another one which converges to $-frac13$. That would also prove that the sequence is not convergent.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 4 hours ago

























                  answered 6 hours ago









                  José Carlos SantosJosé Carlos Santos

                  165k22132235




                  165k22132235























                      1












                      $begingroup$

                      You have a much simpler way to write that. Indeed, suppose that $a_n rightarrow l$. Then
                      $$frac{3n}{n+1}a_n rightarrow 3l$$



                      But $frac{3n}{n+1}a_n = (-1)^n$, therefore it does not converge.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        You have a much simpler way to write that. Indeed, suppose that $a_n rightarrow l$. Then
                        $$frac{3n}{n+1}a_n rightarrow 3l$$



                        But $frac{3n}{n+1}a_n = (-1)^n$, therefore it does not converge.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          You have a much simpler way to write that. Indeed, suppose that $a_n rightarrow l$. Then
                          $$frac{3n}{n+1}a_n rightarrow 3l$$



                          But $frac{3n}{n+1}a_n = (-1)^n$, therefore it does not converge.






                          share|cite|improve this answer









                          $endgroup$



                          You have a much simpler way to write that. Indeed, suppose that $a_n rightarrow l$. Then
                          $$frac{3n}{n+1}a_n rightarrow 3l$$



                          But $frac{3n}{n+1}a_n = (-1)^n$, therefore it does not converge.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 6 hours ago









                          TheSilverDoeTheSilverDoe

                          2,973112




                          2,973112






























                              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%2f3138718%2fverification-of-proof-of-nonconvergence%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

                              Nidaros erkebispedøme

                              Image
                              For det lutherske bispedømet, sjå Nidaros bispedøme. Nidaros erkebispedøme var eit katolsk erkebispedøme med sete i Nidaros. Det vart oppretta av kong Olav Kyrre mellom 1070 og 1080 som bispedøme. Bispedømet vart opphøgd til erkebispedøme i 1153. Erkebispedømet vart innstifta av pave Anastasius IV i 1154, og var i funksjon fram til reformasjonen i 1537. Erkebiskopen sitt hovudsete var Erkebispegarden, rett ved domkyrkja Nidarosdomen. Frå 1152 til 1154 var Nicholas Breakspear i Skandinavia som paveleg utsending, og organiserte opprettinga av erkesetet. Innhaldsliste 1 Utbreiing og jurisdiksjon 2 Overhyrdar 2.1 Lydbiskopar 3 Biskopar 3.1 Biskopar 3.2 Erkebiskopar Utbreiing og jurisdiksjon | Kart over den nordiske kyrkjeprovinsen (1153-1387) Erkebispedømet omfatta i tillegg til Noreg øyane i vest i Noregsveldet. Nidaros var eit metropolitanerkebispedøme med fylgjande underliggjande bispedøme suffraganar: Bj...
                              Read more

                              Birsay

                              Image
                              Tettstaden Birsay på Orknøyane. Ruinane av jarleplasset om lag midt i biletet. (Fotografi frå 2012) Utsikt mot Birsay og Brough of Brisay (fotografi frå 2009) Birsay er ein tettstad i eit sokn (parish) med same namn, nordvest på øya Mainland [Note 1] på Orknøyane, om lag 31 km frå Kirkwall. Staden er omtala fleire stader i Orknøyingasoga [1] , der han er kalla «Byrgesherad» (norrønt: Birgisherað ). Namnet skriv seg truleg frå elvenamnet Birgisá [2] . Innhaldsliste 1 Den gamle hovudstaden 2 Bispesetet 3 Jarlepalasset 4 Kjelder og utfyllande opplysningar 4.1 Notar 4.2 Fotnotar 4.3 Bakgrunnsstoff Den gamle hovudstaden | Hovudartikkelen for dette emnet er Torfinn Sigurdsson . Soknet er i dag i hovudsak ei jordbrukskommune men er mest kjend for dei mange kulturminna i området. Tettstaden Birsay blir gjerne omtala som «den gamle hovudstaden på Orknøyane» [3] , og ein viser d...
                              Read more

                              Was Woodrow Wilson really a Liberal?Was World War I a war of liberals against authoritarians?Founding Fathers...

                              Image
                              Do VLANs within a subnet need to have their own subnet for router on a stick? What are the differences between the usage of 'it' and 'they'? Dragon forelimb placement How do we improve the relationship with a client software team that performs poorly and is becoming less collaborative? What typically incentivizes a professor to change jobs to a lower ranking university? How to write a macro that is braces sensitive? Modeling an IPv4 Address Writing rule stating superpower from different root cause is bad writing Can I ask the recruiters in my resume to put the reason why I am rejected? Approximately how much travel t...
                              Read more