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












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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







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    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






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      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
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          $begingroup$

          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











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              3 Answers
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              3 Answers
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              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


















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