For a sequence $(a_n)$ of real numbers, $sum_{n=1}^infty |a_{n+1}-a_n|$ converges implies...












1












$begingroup$


I know that when the series $sum_{n=1}^infty |a_{n+1}-a_n|$ converges, then we have $|a_{n+1}-a_n|rightarrow 0$

So by using this I was going to prove that the sequence $a_n$ is Cauchy. But couldn't come up with a correct way.






real-analysis sequences-and-series convergence cauchy-sequences







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I know that when the series $sum_{n=1}^infty |a_{n+1}-a_n|$ converges, then we have $|a_{n+1}-a_n|rightarrow 0$

    So by using this I was going to prove that the sequence $a_n$ is Cauchy. But couldn't come up with a correct way.






    real-analysis sequences-and-series convergence cauchy-sequences







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I know that when the series $sum_{n=1}^infty |a_{n+1}-a_n|$ converges, then we have $|a_{n+1}-a_n|rightarrow 0$

      So by using this I was going to prove that the sequence $a_n$ is Cauchy. But couldn't come up with a correct way.






      real-analysis sequences-and-series convergence cauchy-sequences







      share|cite|improve this question











      $endgroup$




      I know that when the series $sum_{n=1}^infty |a_{n+1}-a_n|$ converges, then we have $|a_{n+1}-a_n|rightarrow 0$

      So by using this I was going to prove that the sequence $a_n$ is Cauchy. But couldn't come up with a correct way.






      real-analysis sequences-and-series convergence cauchy-sequences




      real-analysis sequences-and-series convergence cauchy-sequences






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 5 at 20:54









      Davide Giraudo

      127k16154268




      127k16154268










      asked Jan 5 at 20:31









      DD90DD90

      2648




      2648






















          2 Answers
          2






          active

          oldest

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          4












          $begingroup$

          Hint: For every $n geq 0$, $p geq 0$, $$|a_{n+p}-a_n| leq sum_{k=n}^{infty}{|a_{k+1}-a_k|}.$$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Can you please explain more.
            $endgroup$
            – DD90
            Jan 5 at 20:50










          • $begingroup$
            Triangle inequality.
            $endgroup$
            – Math1000
            Jan 5 at 20:51










          • $begingroup$
            Yes. But how does this leads to the convergence of the sequence?
            $endgroup$
            – DD90
            Jan 5 at 20:51












          • $begingroup$
            This proves $(a_n)$ to be Cauchy.
            $endgroup$
            – Mindlack
            Jan 5 at 20:52










          • $begingroup$
            Seems like I'm missing a theory? I'm sorry but still don't get it.. Why does the above inequality says that $forall epsilon >0, exists Ninmathbb{N}$ such that for $p>0$ and $n>N$, $|a_{n+p}-a_n|<epsilon$ ?
            $endgroup$
            – DD90
            Jan 5 at 20:56





















          0












          $begingroup$

          $S_n:=sum_{k=1}^{n}|a_{k+1}-a_{k}|.$



          Since $S_n$ converges , $S_n$ is Cauchy.



          $epsilon >0$ given, there exists a $n_0$ s.t. for $m ge n ge n_0$



          $|S_m-S_n| =|sum_{k=n+1}^{m}|a_{k+1}-a_{k}||$



          $=sum_{k=n+1}^{m}|a_{k+1}-a_k| < epsilon.$



          Show that $(a_n)_{n in mathbb{N}}$ is Cauchy.



          For $m ge n ge (n_0 +1)$ we have:



          $|a_m-a_n| le |a_m-a_{m-1}| +..$



          $|a_{m-1}-a_{m-2}|+.......|a_{n+1}-a_n| $



          $=S_{m-1}-S_{n-1} lt epsilon.$



          Hence Cauchy, hence convergent.






          share|cite|improve this answer









          $endgroup$













            Your Answer





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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            Hint: For every $n geq 0$, $p geq 0$, $$|a_{n+p}-a_n| leq sum_{k=n}^{infty}{|a_{k+1}-a_k|}.$$






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Can you please explain more.
              $endgroup$
              – DD90
              Jan 5 at 20:50










            • $begingroup$
              Triangle inequality.
              $endgroup$
              – Math1000
              Jan 5 at 20:51










            • $begingroup$
              Yes. But how does this leads to the convergence of the sequence?
              $endgroup$
              – DD90
              Jan 5 at 20:51












            • $begingroup$
              This proves $(a_n)$ to be Cauchy.
              $endgroup$
              – Mindlack
              Jan 5 at 20:52










            • $begingroup$
              Seems like I'm missing a theory? I'm sorry but still don't get it.. Why does the above inequality says that $forall epsilon >0, exists Ninmathbb{N}$ such that for $p>0$ and $n>N$, $|a_{n+p}-a_n|<epsilon$ ?
              $endgroup$
              – DD90
              Jan 5 at 20:56


















            4












            $begingroup$

            Hint: For every $n geq 0$, $p geq 0$, $$|a_{n+p}-a_n| leq sum_{k=n}^{infty}{|a_{k+1}-a_k|}.$$






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Can you please explain more.
              $endgroup$
              – DD90
              Jan 5 at 20:50










            • $begingroup$
              Triangle inequality.
              $endgroup$
              – Math1000
              Jan 5 at 20:51










            • $begingroup$
              Yes. But how does this leads to the convergence of the sequence?
              $endgroup$
              – DD90
              Jan 5 at 20:51












            • $begingroup$
              This proves $(a_n)$ to be Cauchy.
              $endgroup$
              – Mindlack
              Jan 5 at 20:52










            • $begingroup$
              Seems like I'm missing a theory? I'm sorry but still don't get it.. Why does the above inequality says that $forall epsilon >0, exists Ninmathbb{N}$ such that for $p>0$ and $n>N$, $|a_{n+p}-a_n|<epsilon$ ?
              $endgroup$
              – DD90
              Jan 5 at 20:56
















            4












            4








            4





            $begingroup$

            Hint: For every $n geq 0$, $p geq 0$, $$|a_{n+p}-a_n| leq sum_{k=n}^{infty}{|a_{k+1}-a_k|}.$$






            share|cite|improve this answer









            $endgroup$



            Hint: For every $n geq 0$, $p geq 0$, $$|a_{n+p}-a_n| leq sum_{k=n}^{infty}{|a_{k+1}-a_k|}.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 5 at 20:42









            MindlackMindlack

            4,885210




            4,885210








            • 1




              $begingroup$
              Can you please explain more.
              $endgroup$
              – DD90
              Jan 5 at 20:50










            • $begingroup$
              Triangle inequality.
              $endgroup$
              – Math1000
              Jan 5 at 20:51










            • $begingroup$
              Yes. But how does this leads to the convergence of the sequence?
              $endgroup$
              – DD90
              Jan 5 at 20:51












            • $begingroup$
              This proves $(a_n)$ to be Cauchy.
              $endgroup$
              – Mindlack
              Jan 5 at 20:52










            • $begingroup$
              Seems like I'm missing a theory? I'm sorry but still don't get it.. Why does the above inequality says that $forall epsilon >0, exists Ninmathbb{N}$ such that for $p>0$ and $n>N$, $|a_{n+p}-a_n|<epsilon$ ?
              $endgroup$
              – DD90
              Jan 5 at 20:56
















            • 1




              $begingroup$
              Can you please explain more.
              $endgroup$
              – DD90
              Jan 5 at 20:50










            • $begingroup$
              Triangle inequality.
              $endgroup$
              – Math1000
              Jan 5 at 20:51










            • $begingroup$
              Yes. But how does this leads to the convergence of the sequence?
              $endgroup$
              – DD90
              Jan 5 at 20:51












            • $begingroup$
              This proves $(a_n)$ to be Cauchy.
              $endgroup$
              – Mindlack
              Jan 5 at 20:52










            • $begingroup$
              Seems like I'm missing a theory? I'm sorry but still don't get it.. Why does the above inequality says that $forall epsilon >0, exists Ninmathbb{N}$ such that for $p>0$ and $n>N$, $|a_{n+p}-a_n|<epsilon$ ?
              $endgroup$
              – DD90
              Jan 5 at 20:56










            1




            1




            $begingroup$
            Can you please explain more.
            $endgroup$
            – DD90
            Jan 5 at 20:50




            $begingroup$
            Can you please explain more.
            $endgroup$
            – DD90
            Jan 5 at 20:50












            $begingroup$
            Triangle inequality.
            $endgroup$
            – Math1000
            Jan 5 at 20:51




            $begingroup$
            Triangle inequality.
            $endgroup$
            – Math1000
            Jan 5 at 20:51












            $begingroup$
            Yes. But how does this leads to the convergence of the sequence?
            $endgroup$
            – DD90
            Jan 5 at 20:51






            $begingroup$
            Yes. But how does this leads to the convergence of the sequence?
            $endgroup$
            – DD90
            Jan 5 at 20:51














            $begingroup$
            This proves $(a_n)$ to be Cauchy.
            $endgroup$
            – Mindlack
            Jan 5 at 20:52




            $begingroup$
            This proves $(a_n)$ to be Cauchy.
            $endgroup$
            – Mindlack
            Jan 5 at 20:52












            $begingroup$
            Seems like I'm missing a theory? I'm sorry but still don't get it.. Why does the above inequality says that $forall epsilon >0, exists Ninmathbb{N}$ such that for $p>0$ and $n>N$, $|a_{n+p}-a_n|<epsilon$ ?
            $endgroup$
            – DD90
            Jan 5 at 20:56






            $begingroup$
            Seems like I'm missing a theory? I'm sorry but still don't get it.. Why does the above inequality says that $forall epsilon >0, exists Ninmathbb{N}$ such that for $p>0$ and $n>N$, $|a_{n+p}-a_n|<epsilon$ ?
            $endgroup$
            – DD90
            Jan 5 at 20:56













            0












            $begingroup$

            $S_n:=sum_{k=1}^{n}|a_{k+1}-a_{k}|.$



            Since $S_n$ converges , $S_n$ is Cauchy.



            $epsilon >0$ given, there exists a $n_0$ s.t. for $m ge n ge n_0$



            $|S_m-S_n| =|sum_{k=n+1}^{m}|a_{k+1}-a_{k}||$



            $=sum_{k=n+1}^{m}|a_{k+1}-a_k| < epsilon.$



            Show that $(a_n)_{n in mathbb{N}}$ is Cauchy.



            For $m ge n ge (n_0 +1)$ we have:



            $|a_m-a_n| le |a_m-a_{m-1}| +..$



            $|a_{m-1}-a_{m-2}|+.......|a_{n+1}-a_n| $



            $=S_{m-1}-S_{n-1} lt epsilon.$



            Hence Cauchy, hence convergent.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              $S_n:=sum_{k=1}^{n}|a_{k+1}-a_{k}|.$



              Since $S_n$ converges , $S_n$ is Cauchy.



              $epsilon >0$ given, there exists a $n_0$ s.t. for $m ge n ge n_0$



              $|S_m-S_n| =|sum_{k=n+1}^{m}|a_{k+1}-a_{k}||$



              $=sum_{k=n+1}^{m}|a_{k+1}-a_k| < epsilon.$



              Show that $(a_n)_{n in mathbb{N}}$ is Cauchy.



              For $m ge n ge (n_0 +1)$ we have:



              $|a_m-a_n| le |a_m-a_{m-1}| +..$



              $|a_{m-1}-a_{m-2}|+.......|a_{n+1}-a_n| $



              $=S_{m-1}-S_{n-1} lt epsilon.$



              Hence Cauchy, hence convergent.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                $S_n:=sum_{k=1}^{n}|a_{k+1}-a_{k}|.$



                Since $S_n$ converges , $S_n$ is Cauchy.



                $epsilon >0$ given, there exists a $n_0$ s.t. for $m ge n ge n_0$



                $|S_m-S_n| =|sum_{k=n+1}^{m}|a_{k+1}-a_{k}||$



                $=sum_{k=n+1}^{m}|a_{k+1}-a_k| < epsilon.$



                Show that $(a_n)_{n in mathbb{N}}$ is Cauchy.



                For $m ge n ge (n_0 +1)$ we have:



                $|a_m-a_n| le |a_m-a_{m-1}| +..$



                $|a_{m-1}-a_{m-2}|+.......|a_{n+1}-a_n| $



                $=S_{m-1}-S_{n-1} lt epsilon.$



                Hence Cauchy, hence convergent.






                share|cite|improve this answer









                $endgroup$



                $S_n:=sum_{k=1}^{n}|a_{k+1}-a_{k}|.$



                Since $S_n$ converges , $S_n$ is Cauchy.



                $epsilon >0$ given, there exists a $n_0$ s.t. for $m ge n ge n_0$



                $|S_m-S_n| =|sum_{k=n+1}^{m}|a_{k+1}-a_{k}||$



                $=sum_{k=n+1}^{m}|a_{k+1}-a_k| < epsilon.$



                Show that $(a_n)_{n in mathbb{N}}$ is Cauchy.



                For $m ge n ge (n_0 +1)$ we have:



                $|a_m-a_n| le |a_m-a_{m-1}| +..$



                $|a_{m-1}-a_{m-2}|+.......|a_{n+1}-a_n| $



                $=S_{m-1}-S_{n-1} lt epsilon.$



                Hence Cauchy, hence convergent.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 5 at 22:28









                Peter SzilasPeter Szilas

                11.4k2822




                11.4k2822






























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