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












1














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







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    1














    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



























      1












      1








      1







      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















      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




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      edited 2 days ago









      Davide Giraudo

      125k16150261




      125k16150261










      asked 2 days ago









      DD90DD90

      2468




      2468






















          2 Answers
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          4














          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

















          • 1




            Can you please explain more.
            – DD90
            2 days ago










          • Triangle inequality.
            – Math1000
            2 days ago










          • Yes. But how does this leads to the convergence of the sequence?
            – DD90
            2 days ago












          • This proves $(a_n)$ to be Cauchy.
            – Mindlack
            2 days ago










          • 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 $n>N$, $|a_{n+p}-a_n|<epsilon$ ?
            – DD90
            2 days ago





















          0














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



          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





















            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4














            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

















            • 1




              Can you please explain more.
              – DD90
              2 days ago










            • Triangle inequality.
              – Math1000
              2 days ago










            • Yes. But how does this leads to the convergence of the sequence?
              – DD90
              2 days ago












            • This proves $(a_n)$ to be Cauchy.
              – Mindlack
              2 days ago










            • 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 $n>N$, $|a_{n+p}-a_n|<epsilon$ ?
              – DD90
              2 days ago


















            4














            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

















            • 1




              Can you please explain more.
              – DD90
              2 days ago










            • Triangle inequality.
              – Math1000
              2 days ago










            • Yes. But how does this leads to the convergence of the sequence?
              – DD90
              2 days ago












            • This proves $(a_n)$ to be Cauchy.
              – Mindlack
              2 days ago










            • 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 $n>N$, $|a_{n+p}-a_n|<epsilon$ ?
              – DD90
              2 days ago
















            4












            4








            4






            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












            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 2 days ago









            MindlackMindlack

            2,06217




            2,06217








            • 1




              Can you please explain more.
              – DD90
              2 days ago










            • Triangle inequality.
              – Math1000
              2 days ago










            • Yes. But how does this leads to the convergence of the sequence?
              – DD90
              2 days ago












            • This proves $(a_n)$ to be Cauchy.
              – Mindlack
              2 days ago










            • 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 $n>N$, $|a_{n+p}-a_n|<epsilon$ ?
              – DD90
              2 days ago
















            • 1




              Can you please explain more.
              – DD90
              2 days ago










            • Triangle inequality.
              – Math1000
              2 days ago










            • Yes. But how does this leads to the convergence of the sequence?
              – DD90
              2 days ago












            • This proves $(a_n)$ to be Cauchy.
              – Mindlack
              2 days ago










            • 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 $n>N$, $|a_{n+p}-a_n|<epsilon$ ?
              – DD90
              2 days ago










            1




            1




            Can you please explain more.
            – DD90
            2 days ago




            Can you please explain more.
            – DD90
            2 days ago












            Triangle inequality.
            – Math1000
            2 days ago




            Triangle inequality.
            – Math1000
            2 days ago












            Yes. But how does this leads to the convergence of the sequence?
            – DD90
            2 days ago






            Yes. But how does this leads to the convergence of the sequence?
            – DD90
            2 days ago














            This proves $(a_n)$ to be Cauchy.
            – Mindlack
            2 days ago




            This proves $(a_n)$ to be Cauchy.
            – Mindlack
            2 days ago












            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 $n>N$, $|a_{n+p}-a_n|<epsilon$ ?
            – DD90
            2 days ago






            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 $n>N$, $|a_{n+p}-a_n|<epsilon$ ?
            – DD90
            2 days ago













            0














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



            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


























              0














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



              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
























                0












                0








                0






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



                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












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



                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 2 days ago









                Peter SzilasPeter Szilas

                10.8k2720




                10.8k2720






























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