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Proving that a sequence of reciprocated integers is not Cauchy [on hold]


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$begingroup$


Let ${a_n}_n$ be the sequence given by $a_n = 1 + frac 12 + cdots + frac 1n.$ Prove the sequence is not Cauchy.



I don't really have any idea where to start with this problem. I know the sequence diverges, but I can't think of any way to prove this one. Appreciate any help, thank you.






real-analysis cauchy-sequences







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put on hold as off-topic by RRL, YiFan, Leucippus, Eevee Trainer, Xander Henderson 45 mins ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, YiFan, Leucippus, Eevee Trainer, Xander Henderson

If this question can be reworded to fit the rules in the help center, please edit the question.





















    -1












    $begingroup$


    Let ${a_n}_n$ be the sequence given by $a_n = 1 + frac 12 + cdots + frac 1n.$ Prove the sequence is not Cauchy.



    I don't really have any idea where to start with this problem. I know the sequence diverges, but I can't think of any way to prove this one. Appreciate any help, thank you.






    real-analysis cauchy-sequences







    share|cite|improve this question









    $endgroup$



    put on hold as off-topic by RRL, YiFan, Leucippus, Eevee Trainer, Xander Henderson 45 mins ago


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, YiFan, Leucippus, Eevee Trainer, Xander Henderson

    If this question can be reworded to fit the rules in the help center, please edit the question.



















      -1












      -1








      -1





      $begingroup$


      Let ${a_n}_n$ be the sequence given by $a_n = 1 + frac 12 + cdots + frac 1n.$ Prove the sequence is not Cauchy.



      I don't really have any idea where to start with this problem. I know the sequence diverges, but I can't think of any way to prove this one. Appreciate any help, thank you.






      real-analysis cauchy-sequences







      share|cite|improve this question









      $endgroup$




      Let ${a_n}_n$ be the sequence given by $a_n = 1 + frac 12 + cdots + frac 1n.$ Prove the sequence is not Cauchy.



      I don't really have any idea where to start with this problem. I know the sequence diverges, but I can't think of any way to prove this one. Appreciate any help, thank you.






      real-analysis cauchy-sequences




      real-analysis cauchy-sequences






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 20 hours ago









      B RetnikB Retnik

      534




      534




      put on hold as off-topic by RRL, YiFan, Leucippus, Eevee Trainer, Xander Henderson 45 mins ago


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, YiFan, Leucippus, Eevee Trainer, Xander Henderson

      If this question can be reworded to fit the rules in the help center, please edit the question.







      put on hold as off-topic by RRL, YiFan, Leucippus, Eevee Trainer, Xander Henderson 45 mins ago


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, YiFan, Leucippus, Eevee Trainer, Xander Henderson

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          2 Answers
          2






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          3












          $begingroup$

          We have $a_{2n}=a_n+frac{1}{n+1}+...+frac{1}{2n} ge a_n+n cdot frac{1}{2n}=a_n+frac{1}{2}.$



          Hence $|a_{2n}-a_n|=a_{2n}-a_n ge frac{1}{2}$ for all $n$.



          Can you take it from here ?






          share|cite|improve this answer









          $endgroup$





















            4












            $begingroup$

            $frac 1 n+frac 1 {n+1} +cdots+frac 1 {k_n} geq frac {k_n-n+1} {k_n}$ if $k_n >n$. We can choose $k_n$ such that $frac {k_n-n+1} {k_n}$ does not tend to $0$, e.g. $k_n=n^{2}$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Why is $frac 1n + frac 1{n+1} + cdots + frac 1{k_n} ge frac {k_n-n+1}{k_n}$?
              $endgroup$
              – B Retnik
              19 hours ago






            • 2




              $begingroup$
              Because there are $k_n-n+1$ terms and each term is $geq frac 1 {k_n}$.
              $endgroup$
              – Kavi Rama Murthy
              19 hours ago


















            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            We have $a_{2n}=a_n+frac{1}{n+1}+...+frac{1}{2n} ge a_n+n cdot frac{1}{2n}=a_n+frac{1}{2}.$



            Hence $|a_{2n}-a_n|=a_{2n}-a_n ge frac{1}{2}$ for all $n$.



            Can you take it from here ?






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              We have $a_{2n}=a_n+frac{1}{n+1}+...+frac{1}{2n} ge a_n+n cdot frac{1}{2n}=a_n+frac{1}{2}.$



              Hence $|a_{2n}-a_n|=a_{2n}-a_n ge frac{1}{2}$ for all $n$.



              Can you take it from here ?






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                We have $a_{2n}=a_n+frac{1}{n+1}+...+frac{1}{2n} ge a_n+n cdot frac{1}{2n}=a_n+frac{1}{2}.$



                Hence $|a_{2n}-a_n|=a_{2n}-a_n ge frac{1}{2}$ for all $n$.



                Can you take it from here ?






                share|cite|improve this answer









                $endgroup$



                We have $a_{2n}=a_n+frac{1}{n+1}+...+frac{1}{2n} ge a_n+n cdot frac{1}{2n}=a_n+frac{1}{2}.$



                Hence $|a_{2n}-a_n|=a_{2n}-a_n ge frac{1}{2}$ for all $n$.



                Can you take it from here ?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 19 hours ago









                FredFred

                47.7k1849




                47.7k1849























                    4












                    $begingroup$

                    $frac 1 n+frac 1 {n+1} +cdots+frac 1 {k_n} geq frac {k_n-n+1} {k_n}$ if $k_n >n$. We can choose $k_n$ such that $frac {k_n-n+1} {k_n}$ does not tend to $0$, e.g. $k_n=n^{2}$.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Why is $frac 1n + frac 1{n+1} + cdots + frac 1{k_n} ge frac {k_n-n+1}{k_n}$?
                      $endgroup$
                      – B Retnik
                      19 hours ago






                    • 2




                      $begingroup$
                      Because there are $k_n-n+1$ terms and each term is $geq frac 1 {k_n}$.
                      $endgroup$
                      – Kavi Rama Murthy
                      19 hours ago
















                    4












                    $begingroup$

                    $frac 1 n+frac 1 {n+1} +cdots+frac 1 {k_n} geq frac {k_n-n+1} {k_n}$ if $k_n >n$. We can choose $k_n$ such that $frac {k_n-n+1} {k_n}$ does not tend to $0$, e.g. $k_n=n^{2}$.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Why is $frac 1n + frac 1{n+1} + cdots + frac 1{k_n} ge frac {k_n-n+1}{k_n}$?
                      $endgroup$
                      – B Retnik
                      19 hours ago






                    • 2




                      $begingroup$
                      Because there are $k_n-n+1$ terms and each term is $geq frac 1 {k_n}$.
                      $endgroup$
                      – Kavi Rama Murthy
                      19 hours ago














                    4












                    4








                    4





                    $begingroup$

                    $frac 1 n+frac 1 {n+1} +cdots+frac 1 {k_n} geq frac {k_n-n+1} {k_n}$ if $k_n >n$. We can choose $k_n$ such that $frac {k_n-n+1} {k_n}$ does not tend to $0$, e.g. $k_n=n^{2}$.






                    share|cite|improve this answer









                    $endgroup$



                    $frac 1 n+frac 1 {n+1} +cdots+frac 1 {k_n} geq frac {k_n-n+1} {k_n}$ if $k_n >n$. We can choose $k_n$ such that $frac {k_n-n+1} {k_n}$ does not tend to $0$, e.g. $k_n=n^{2}$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 20 hours ago









                    Kavi Rama MurthyKavi Rama Murthy

                    65.4k42766




                    65.4k42766












                    • $begingroup$
                      Why is $frac 1n + frac 1{n+1} + cdots + frac 1{k_n} ge frac {k_n-n+1}{k_n}$?
                      $endgroup$
                      – B Retnik
                      19 hours ago






                    • 2




                      $begingroup$
                      Because there are $k_n-n+1$ terms and each term is $geq frac 1 {k_n}$.
                      $endgroup$
                      – Kavi Rama Murthy
                      19 hours ago


















                    • $begingroup$
                      Why is $frac 1n + frac 1{n+1} + cdots + frac 1{k_n} ge frac {k_n-n+1}{k_n}$?
                      $endgroup$
                      – B Retnik
                      19 hours ago






                    • 2




                      $begingroup$
                      Because there are $k_n-n+1$ terms and each term is $geq frac 1 {k_n}$.
                      $endgroup$
                      – Kavi Rama Murthy
                      19 hours ago
















                    $begingroup$
                    Why is $frac 1n + frac 1{n+1} + cdots + frac 1{k_n} ge frac {k_n-n+1}{k_n}$?
                    $endgroup$
                    – B Retnik
                    19 hours ago




                    $begingroup$
                    Why is $frac 1n + frac 1{n+1} + cdots + frac 1{k_n} ge frac {k_n-n+1}{k_n}$?
                    $endgroup$
                    – B Retnik
                    19 hours ago




                    2




                    2




                    $begingroup$
                    Because there are $k_n-n+1$ terms and each term is $geq frac 1 {k_n}$.
                    $endgroup$
                    – Kavi Rama Murthy
                    19 hours ago




                    $begingroup$
                    Because there are $k_n-n+1$ terms and each term is $geq frac 1 {k_n}$.
                    $endgroup$
                    – Kavi Rama Murthy
                    19 hours ago



                    -

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