If $lim limits_{n to infty } (a_{n+1}-a_n)=0$ and $(a_{4n})_{nge1}$ converges decide whether or not $(a_n)$...












0












$begingroup$


If $lim limits_{n to infty } (a_{n+1}-a_n)=0$ and $(a_{4n})_{nge1}$ converges decide whether or not $(a_n)$ converges.



My attempt:



I think it isn't enough to $a_n$ converges.



I know $lim limits_{n to infty }( a_{4n+1}-a_{4n})=0$, and because $(a_{4n+1}−a_{4n})$ is a subsequence of $(a_{n+1}-a_n)$ it also converges to 0, so $lim limits_{n to infty } a_{4n+1}=lim limits_{n to infty } a_{4n} $



But I don't know how how to finish this proof. Could someone give me some hints?






real-analysis sequences-and-series limits-without-lhopital







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    If $lim limits_{n to infty } (a_{n+1}-a_n)=0$ and $(a_{4n})_{nge1}$ converges decide whether or not $(a_n)$ converges.



    My attempt:



    I think it isn't enough to $a_n$ converges.



    I know $lim limits_{n to infty }( a_{4n+1}-a_{4n})=0$, and because $(a_{4n+1}−a_{4n})$ is a subsequence of $(a_{n+1}-a_n)$ it also converges to 0, so $lim limits_{n to infty } a_{4n+1}=lim limits_{n to infty } a_{4n} $



    But I don't know how how to finish this proof. Could someone give me some hints?






    real-analysis sequences-and-series limits-without-lhopital







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      If $lim limits_{n to infty } (a_{n+1}-a_n)=0$ and $(a_{4n})_{nge1}$ converges decide whether or not $(a_n)$ converges.



      My attempt:



      I think it isn't enough to $a_n$ converges.



      I know $lim limits_{n to infty }( a_{4n+1}-a_{4n})=0$, and because $(a_{4n+1}−a_{4n})$ is a subsequence of $(a_{n+1}-a_n)$ it also converges to 0, so $lim limits_{n to infty } a_{4n+1}=lim limits_{n to infty } a_{4n} $



      But I don't know how how to finish this proof. Could someone give me some hints?






      real-analysis sequences-and-series limits-without-lhopital







      share|cite|improve this question











      $endgroup$




      If $lim limits_{n to infty } (a_{n+1}-a_n)=0$ and $(a_{4n})_{nge1}$ converges decide whether or not $(a_n)$ converges.



      My attempt:



      I think it isn't enough to $a_n$ converges.



      I know $lim limits_{n to infty }( a_{4n+1}-a_{4n})=0$, and because $(a_{4n+1}−a_{4n})$ is a subsequence of $(a_{n+1}-a_n)$ it also converges to 0, so $lim limits_{n to infty } a_{4n+1}=lim limits_{n to infty } a_{4n} $



      But I don't know how how to finish this proof. Could someone give me some hints?






      real-analysis sequences-and-series limits-without-lhopital




      real-analysis sequences-and-series limits-without-lhopital






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 30 '18 at 8:33









      Mostafa Ayaz

      15.5k3939




      15.5k3939










      asked Nov 29 '18 at 23:32









      matematicccmatematiccc

      1275




      1275






















          3 Answers
          3






          active

          oldest

          votes


















          4












          $begingroup$

          Write $ell = lim_{ntoinfty} a_{4n}$. For each $epsilon > 0$,




          • Pick $N_1$ such that $n geq N_1$ implies $|a_{n+1} - a_n| < epsilon/5$.

          • Pick $N_2$ such that $4k geq N_2$ implies $|a_{4k} - ell| < epsilon/5$.


          Write $N = max{N_1, N_2}$. Then for each $n geq N$, with $n = 4k+r$ with $r in {0,1,2,3}$,



          $$ |a_n - ell| leq left( sum_{j=r}^{3} |a_{4k+j+1} - a_{4k+j}| right) + |a_{4(k+1)} - ell| < epsilon. $$



          Therefore the $epsilon$-$N$ definition of $a_n to ell$ is established and hence $a_n to ell$.






          share|cite|improve this answer











          $endgroup$





















            2












            $begingroup$

            Hints: let $j,kin mathbb N$ and write $j=4j'+r,k=4k'+s$ with $0leq r,s <4$. Using the fact that $a_{n+1}-a_n to 0$ show that $a_{n+2}-a_n to 0$ and $a_{n+3}-a_n to 0$. Conclude that $a_j-a_{4j'} to 0$ and $a_k-a_{4k'} to 0$. Now can you complete the proof?






            share|cite|improve this answer











            $endgroup$





















              1












              $begingroup$

              By the same technique you used, $(a_{4n+2} - a_{4n+1})$ is a subsequence of $(a_{n+1} - a_n)$, so it has to converge to $0$ as well. Does that help?






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                So $lim limits_{n to infty } a_{4n}=lim limits_{n to infty } a_{4n+1}=lim limits_{n to infty } a_{4n+2}+lim limits_{n to infty } a_{4n+3}$, and from this, it equals $lim limits_{n to infty } a_{n}$ so $ a_n $ converges, true?
                $endgroup$
                – matematiccc
                Nov 29 '18 at 23:45












              • $begingroup$
                Yep (to be precise, you may want to break it down into $epsilon$ and $N$, but that's the basic idea).
                $endgroup$
                – platty
                Nov 29 '18 at 23:46











              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%2f3019396%2fif-lim-limits-n-to-infty-a-n1-a-n-0-and-a-4n-n-ge1-conver%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









              4












              $begingroup$

              Write $ell = lim_{ntoinfty} a_{4n}$. For each $epsilon > 0$,




              • Pick $N_1$ such that $n geq N_1$ implies $|a_{n+1} - a_n| < epsilon/5$.

              • Pick $N_2$ such that $4k geq N_2$ implies $|a_{4k} - ell| < epsilon/5$.


              Write $N = max{N_1, N_2}$. Then for each $n geq N$, with $n = 4k+r$ with $r in {0,1,2,3}$,



              $$ |a_n - ell| leq left( sum_{j=r}^{3} |a_{4k+j+1} - a_{4k+j}| right) + |a_{4(k+1)} - ell| < epsilon. $$



              Therefore the $epsilon$-$N$ definition of $a_n to ell$ is established and hence $a_n to ell$.






              share|cite|improve this answer











              $endgroup$


















                4












                $begingroup$

                Write $ell = lim_{ntoinfty} a_{4n}$. For each $epsilon > 0$,




                • Pick $N_1$ such that $n geq N_1$ implies $|a_{n+1} - a_n| < epsilon/5$.

                • Pick $N_2$ such that $4k geq N_2$ implies $|a_{4k} - ell| < epsilon/5$.


                Write $N = max{N_1, N_2}$. Then for each $n geq N$, with $n = 4k+r$ with $r in {0,1,2,3}$,



                $$ |a_n - ell| leq left( sum_{j=r}^{3} |a_{4k+j+1} - a_{4k+j}| right) + |a_{4(k+1)} - ell| < epsilon. $$



                Therefore the $epsilon$-$N$ definition of $a_n to ell$ is established and hence $a_n to ell$.






                share|cite|improve this answer











                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  Write $ell = lim_{ntoinfty} a_{4n}$. For each $epsilon > 0$,




                  • Pick $N_1$ such that $n geq N_1$ implies $|a_{n+1} - a_n| < epsilon/5$.

                  • Pick $N_2$ such that $4k geq N_2$ implies $|a_{4k} - ell| < epsilon/5$.


                  Write $N = max{N_1, N_2}$. Then for each $n geq N$, with $n = 4k+r$ with $r in {0,1,2,3}$,



                  $$ |a_n - ell| leq left( sum_{j=r}^{3} |a_{4k+j+1} - a_{4k+j}| right) + |a_{4(k+1)} - ell| < epsilon. $$



                  Therefore the $epsilon$-$N$ definition of $a_n to ell$ is established and hence $a_n to ell$.






                  share|cite|improve this answer











                  $endgroup$



                  Write $ell = lim_{ntoinfty} a_{4n}$. For each $epsilon > 0$,




                  • Pick $N_1$ such that $n geq N_1$ implies $|a_{n+1} - a_n| < epsilon/5$.

                  • Pick $N_2$ such that $4k geq N_2$ implies $|a_{4k} - ell| < epsilon/5$.


                  Write $N = max{N_1, N_2}$. Then for each $n geq N$, with $n = 4k+r$ with $r in {0,1,2,3}$,



                  $$ |a_n - ell| leq left( sum_{j=r}^{3} |a_{4k+j+1} - a_{4k+j}| right) + |a_{4(k+1)} - ell| < epsilon. $$



                  Therefore the $epsilon$-$N$ definition of $a_n to ell$ is established and hence $a_n to ell$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 30 '18 at 17:06

























                  answered Nov 30 '18 at 8:53









                  Sangchul LeeSangchul Lee

                  93.3k12167270




                  93.3k12167270























                      2












                      $begingroup$

                      Hints: let $j,kin mathbb N$ and write $j=4j'+r,k=4k'+s$ with $0leq r,s <4$. Using the fact that $a_{n+1}-a_n to 0$ show that $a_{n+2}-a_n to 0$ and $a_{n+3}-a_n to 0$. Conclude that $a_j-a_{4j'} to 0$ and $a_k-a_{4k'} to 0$. Now can you complete the proof?






                      share|cite|improve this answer











                      $endgroup$


















                        2












                        $begingroup$

                        Hints: let $j,kin mathbb N$ and write $j=4j'+r,k=4k'+s$ with $0leq r,s <4$. Using the fact that $a_{n+1}-a_n to 0$ show that $a_{n+2}-a_n to 0$ and $a_{n+3}-a_n to 0$. Conclude that $a_j-a_{4j'} to 0$ and $a_k-a_{4k'} to 0$. Now can you complete the proof?






                        share|cite|improve this answer











                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          Hints: let $j,kin mathbb N$ and write $j=4j'+r,k=4k'+s$ with $0leq r,s <4$. Using the fact that $a_{n+1}-a_n to 0$ show that $a_{n+2}-a_n to 0$ and $a_{n+3}-a_n to 0$. Conclude that $a_j-a_{4j'} to 0$ and $a_k-a_{4k'} to 0$. Now can you complete the proof?






                          share|cite|improve this answer











                          $endgroup$



                          Hints: let $j,kin mathbb N$ and write $j=4j'+r,k=4k'+s$ with $0leq r,s <4$. Using the fact that $a_{n+1}-a_n to 0$ show that $a_{n+2}-a_n to 0$ and $a_{n+3}-a_n to 0$. Conclude that $a_j-a_{4j'} to 0$ and $a_k-a_{4k'} to 0$. Now can you complete the proof?







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Nov 29 '18 at 23:46









                          José Carlos Santos

                          160k22126232




                          160k22126232










                          answered Nov 29 '18 at 23:38









                          Kavi Rama MurthyKavi Rama Murthy

                          58.8k42161




                          58.8k42161























                              1












                              $begingroup$

                              By the same technique you used, $(a_{4n+2} - a_{4n+1})$ is a subsequence of $(a_{n+1} - a_n)$, so it has to converge to $0$ as well. Does that help?






                              share|cite|improve this answer









                              $endgroup$













                              • $begingroup$
                                So $lim limits_{n to infty } a_{4n}=lim limits_{n to infty } a_{4n+1}=lim limits_{n to infty } a_{4n+2}+lim limits_{n to infty } a_{4n+3}$, and from this, it equals $lim limits_{n to infty } a_{n}$ so $ a_n $ converges, true?
                                $endgroup$
                                – matematiccc
                                Nov 29 '18 at 23:45












                              • $begingroup$
                                Yep (to be precise, you may want to break it down into $epsilon$ and $N$, but that's the basic idea).
                                $endgroup$
                                – platty
                                Nov 29 '18 at 23:46
















                              1












                              $begingroup$

                              By the same technique you used, $(a_{4n+2} - a_{4n+1})$ is a subsequence of $(a_{n+1} - a_n)$, so it has to converge to $0$ as well. Does that help?






                              share|cite|improve this answer









                              $endgroup$













                              • $begingroup$
                                So $lim limits_{n to infty } a_{4n}=lim limits_{n to infty } a_{4n+1}=lim limits_{n to infty } a_{4n+2}+lim limits_{n to infty } a_{4n+3}$, and from this, it equals $lim limits_{n to infty } a_{n}$ so $ a_n $ converges, true?
                                $endgroup$
                                – matematiccc
                                Nov 29 '18 at 23:45












                              • $begingroup$
                                Yep (to be precise, you may want to break it down into $epsilon$ and $N$, but that's the basic idea).
                                $endgroup$
                                – platty
                                Nov 29 '18 at 23:46














                              1












                              1








                              1





                              $begingroup$

                              By the same technique you used, $(a_{4n+2} - a_{4n+1})$ is a subsequence of $(a_{n+1} - a_n)$, so it has to converge to $0$ as well. Does that help?






                              share|cite|improve this answer









                              $endgroup$



                              By the same technique you used, $(a_{4n+2} - a_{4n+1})$ is a subsequence of $(a_{n+1} - a_n)$, so it has to converge to $0$ as well. Does that help?







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Nov 29 '18 at 23:36









                              plattyplatty

                              3,370320




                              3,370320












                              • $begingroup$
                                So $lim limits_{n to infty } a_{4n}=lim limits_{n to infty } a_{4n+1}=lim limits_{n to infty } a_{4n+2}+lim limits_{n to infty } a_{4n+3}$, and from this, it equals $lim limits_{n to infty } a_{n}$ so $ a_n $ converges, true?
                                $endgroup$
                                – matematiccc
                                Nov 29 '18 at 23:45












                              • $begingroup$
                                Yep (to be precise, you may want to break it down into $epsilon$ and $N$, but that's the basic idea).
                                $endgroup$
                                – platty
                                Nov 29 '18 at 23:46


















                              • $begingroup$
                                So $lim limits_{n to infty } a_{4n}=lim limits_{n to infty } a_{4n+1}=lim limits_{n to infty } a_{4n+2}+lim limits_{n to infty } a_{4n+3}$, and from this, it equals $lim limits_{n to infty } a_{n}$ so $ a_n $ converges, true?
                                $endgroup$
                                – matematiccc
                                Nov 29 '18 at 23:45












                              • $begingroup$
                                Yep (to be precise, you may want to break it down into $epsilon$ and $N$, but that's the basic idea).
                                $endgroup$
                                – platty
                                Nov 29 '18 at 23:46
















                              $begingroup$
                              So $lim limits_{n to infty } a_{4n}=lim limits_{n to infty } a_{4n+1}=lim limits_{n to infty } a_{4n+2}+lim limits_{n to infty } a_{4n+3}$, and from this, it equals $lim limits_{n to infty } a_{n}$ so $ a_n $ converges, true?
                              $endgroup$
                              – matematiccc
                              Nov 29 '18 at 23:45






                              $begingroup$
                              So $lim limits_{n to infty } a_{4n}=lim limits_{n to infty } a_{4n+1}=lim limits_{n to infty } a_{4n+2}+lim limits_{n to infty } a_{4n+3}$, and from this, it equals $lim limits_{n to infty } a_{n}$ so $ a_n $ converges, true?
                              $endgroup$
                              – matematiccc
                              Nov 29 '18 at 23:45














                              $begingroup$
                              Yep (to be precise, you may want to break it down into $epsilon$ and $N$, but that's the basic idea).
                              $endgroup$
                              – platty
                              Nov 29 '18 at 23:46




                              $begingroup$
                              Yep (to be precise, you may want to break it down into $epsilon$ and $N$, but that's the basic idea).
                              $endgroup$
                              – platty
                              Nov 29 '18 at 23:46


















                              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%2f3019396%2fif-lim-limits-n-to-infty-a-n1-a-n-0-and-a-4n-n-ge1-conver%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

                              Biblatex bibliography style without URLs when DOI exists (in Overleaf with Zotero bibliography)

                              Image
                              2 2 I have a Zotero bibliography in Overleaf. I use biblatex . usepackage[backend=biber,citestyle=authoryear]{biblatex} addbibresource{zoteroALH.bib} Bibliography entries look like: @book{de_saint-gervais_uniformization_2016, title = {Uniformization of {Riemann} {Surfaces}}, isbn = {978-3-03719-145-3}, url = {https://www.ems-ph.org/books/book.php?proj_nr=198&srch=series%7Chem}, urldate = {2019-02-03}, author = {de Saint-Gervais, Paul Henri}, year = {2016}, doi = {10.4171/145} } In the bibliography down the paper the reference shows like this: There is the DOI, great, but also the URL which is unnecessary when the DOI is there. How can I remove the URL when the DOI is present? biblatex bibliographies overleaf zote
                              Read more

                              ComboBox Display Member on multiple fields

                              Image
                              2 I'm trying to set the .DisplayMember property of a ComboBox in C#, but I want to bind it to multiple columns in the .DataSouce . My SQL looks like this: SELECT PersNbr, PersFirstName, PersMiddleName, PersLastName FROM Pers WHERE PersNbr = :persNbr; I'm saving this query in a DataTable so each column selected has it's own column in the Datatable . I want to make the .DisplayMember a combination of PersFirstName + PersMiddleName + PersLastName so their full name appears like this: comboBox.DisplayMemeber = "PersFirstName" + "PersMiddleName" + "PersLastName" I know I can just to this on the query: SELECT PersNbr, (PersFirstName || PersMiddleName || PersLastName) PersName and then just do this: comboBox.DisplayMember = "PersName"; but I
                              Read more

                              Is it possible to collect Nectar points via Trainline?

                              Image
                              2 Since I am not eligible for any Railcard, the only available discount/bonus scheme is Nectar Card for me. GWR, SWR, LNER and Virgin Trains allow to collect Nectar points and I have already linked a card to my GWR account. A friend suggested me to use Trainline App, because it can search for bus tickets either. I had a look at the application, but could not see any option for Nectar. Is there a way to collect Nectar points when I buy tickets via Trainline? uk trains loyalty-programs trainline share | improve this question asked Dec 2 '18 at 15:50 ahmedus 3,224 5 19 49
                              Read more