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

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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?













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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







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    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






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      edited 6 hours ago









      José Carlos Santos

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      asked 6 hours ago









      B RetnikB Retnik

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          $begingroup$

          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











          $endgroup$





















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            $begingroup$

            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









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              $begingroup$

              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






























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