Convergence of $sum_{n=1}^inftyleft(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}right)$ where $a_1=1$ and $a_n=2-frac...












2












$begingroup$


Let $a_1=1 $ and $ a_n=2-frac 1n$ for $ngeq 2$. Then $$sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)$$
converges to?




I started by expanding the sum:



$sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)=bigg(frac{1}{a_1^2}-frac{1}{a_2^2}bigg)+bigg(frac{1}{a_2^2}-frac{1}{a_3^2}bigg)+bigg(frac{1}{a_3^2}-frac{1}{a_4^2}bigg)+ldots$



$=frac{1}{a_1^2}+bigg(-frac{1}{a_2^2}+frac{1}{a_2^2}bigg)+bigg(-frac{1}{a_3^2}+frac{1}{a_3^2}bigg)+bigg(-frac{1}{a_4^2}+frac{1}{a_4^2}bigg)+ldots$



$=frac{1}{a_1^2}$



$=1$




Now, the strange part is that the answer to this question is known to me as $0.75$. Is there something wrong with what I did? How does one arrive at $0.75$ as a solution?



PS: I am aware that infinite sums can behave in strange ways, and sometimes more than one solution may follow logically. Is this one such case?






sequences-and-series convergence







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$endgroup$

















    2












    $begingroup$


    Let $a_1=1 $ and $ a_n=2-frac 1n$ for $ngeq 2$. Then $$sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)$$
    converges to?




    I started by expanding the sum:



    $sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)=bigg(frac{1}{a_1^2}-frac{1}{a_2^2}bigg)+bigg(frac{1}{a_2^2}-frac{1}{a_3^2}bigg)+bigg(frac{1}{a_3^2}-frac{1}{a_4^2}bigg)+ldots$



    $=frac{1}{a_1^2}+bigg(-frac{1}{a_2^2}+frac{1}{a_2^2}bigg)+bigg(-frac{1}{a_3^2}+frac{1}{a_3^2}bigg)+bigg(-frac{1}{a_4^2}+frac{1}{a_4^2}bigg)+ldots$



    $=frac{1}{a_1^2}$



    $=1$




    Now, the strange part is that the answer to this question is known to me as $0.75$. Is there something wrong with what I did? How does one arrive at $0.75$ as a solution?



    PS: I am aware that infinite sums can behave in strange ways, and sometimes more than one solution may follow logically. Is this one such case?






    sequences-and-series convergence







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Let $a_1=1 $ and $ a_n=2-frac 1n$ for $ngeq 2$. Then $$sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)$$
      converges to?




      I started by expanding the sum:



      $sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)=bigg(frac{1}{a_1^2}-frac{1}{a_2^2}bigg)+bigg(frac{1}{a_2^2}-frac{1}{a_3^2}bigg)+bigg(frac{1}{a_3^2}-frac{1}{a_4^2}bigg)+ldots$



      $=frac{1}{a_1^2}+bigg(-frac{1}{a_2^2}+frac{1}{a_2^2}bigg)+bigg(-frac{1}{a_3^2}+frac{1}{a_3^2}bigg)+bigg(-frac{1}{a_4^2}+frac{1}{a_4^2}bigg)+ldots$



      $=frac{1}{a_1^2}$



      $=1$




      Now, the strange part is that the answer to this question is known to me as $0.75$. Is there something wrong with what I did? How does one arrive at $0.75$ as a solution?



      PS: I am aware that infinite sums can behave in strange ways, and sometimes more than one solution may follow logically. Is this one such case?






      sequences-and-series convergence







      share|cite|improve this question











      $endgroup$




      Let $a_1=1 $ and $ a_n=2-frac 1n$ for $ngeq 2$. Then $$sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)$$
      converges to?




      I started by expanding the sum:



      $sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)=bigg(frac{1}{a_1^2}-frac{1}{a_2^2}bigg)+bigg(frac{1}{a_2^2}-frac{1}{a_3^2}bigg)+bigg(frac{1}{a_3^2}-frac{1}{a_4^2}bigg)+ldots$



      $=frac{1}{a_1^2}+bigg(-frac{1}{a_2^2}+frac{1}{a_2^2}bigg)+bigg(-frac{1}{a_3^2}+frac{1}{a_3^2}bigg)+bigg(-frac{1}{a_4^2}+frac{1}{a_4^2}bigg)+ldots$



      $=frac{1}{a_1^2}$



      $=1$




      Now, the strange part is that the answer to this question is known to me as $0.75$. Is there something wrong with what I did? How does one arrive at $0.75$ as a solution?



      PS: I am aware that infinite sums can behave in strange ways, and sometimes more than one solution may follow logically. Is this one such case?






      sequences-and-series convergence




      sequences-and-series convergence






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 22 hours ago









      Asaf Karagila♦

      303k32429761




      303k32429761










      asked yesterday









      s0ulr3aper07s0ulr3aper07

      3119




      3119






















          3 Answers
          3






          active

          oldest

          votes


















          7












          $begingroup$

          Let $$S_N=sum_{n=1}^Nleft(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}right)=frac{1}{a_1^2}-frac{1}{a_{N+1}^2}.$$
          Since $a_1=1$ and $a_nto 2$, we have
          $$lim_{Ntoinfty}S_N=1-frac{1}{2^2}=frac{3}{4}.$$






          share|cite|improve this answer









          $endgroup$





















            4












            $begingroup$


            Infinite sums can behave in strange ways




            True!




            Sometimes more than one solution may follow logically




            Definitely not true. There is one answer in every case (as long as you allow "the sum diverges" as an answer). Sure, there are subtle errors which may make it appear that a sum has more than one value, but that's exactly what they are - errors.



            To avoid these errors, don't manipulate infinite sums unless you have a conclusive proof that the manipulation is justified. If you don't, go back to the definition. In your case this says
            $$eqalign{sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)
            &=lim_{Ntoinfty}sum_{n=1}^Nbigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)cr
            &=lim_{Ntoinfty}biggl(frac{1}{a_1^2}-frac{1}{a_{N+1}^2}biggr) .cr}$$
            See if you can finish the job from here.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I stated that more than one solution may follow logically because of Grandi's series where 0, 1, and half are possible answers. Although, I suppose you could simply call the series divergent and completely avoid the controversy altogether.
              $endgroup$
              – s0ulr3aper07
              yesterday






            • 2




              $begingroup$
              Using standard definitions, Grandi's series is divergent and that's that. Using Cesàro summation you get $frac12$, but this does not mean the series actually converges to $frac12$. As far as I am aware, there has been absolutely no serious controversy on this question for well over a century. Well stated in Wikipedia: "the above manipulations [giving various 'sums'] do not consider what the sum of a series actually means".
              $endgroup$
              – David
              yesterday





















            3












            $begingroup$

            $a_n$ does not approach zero,
            so the end term can not
            be disregarded.



            In other words,
            the sum up to $n$ is
            $frac1{a_1^2}
            -frac1{a_n^2}
            to 1-frac14
            =frac34
            $            .






            share|cite|improve this answer









            $endgroup$













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              3 Answers
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              3 Answers
              3






              active

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              active

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              7












              $begingroup$

              Let $$S_N=sum_{n=1}^Nleft(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}right)=frac{1}{a_1^2}-frac{1}{a_{N+1}^2}.$$
              Since $a_1=1$ and $a_nto 2$, we have
              $$lim_{Ntoinfty}S_N=1-frac{1}{2^2}=frac{3}{4}.$$






              share|cite|improve this answer









              $endgroup$


















                7












                $begingroup$

                Let $$S_N=sum_{n=1}^Nleft(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}right)=frac{1}{a_1^2}-frac{1}{a_{N+1}^2}.$$
                Since $a_1=1$ and $a_nto 2$, we have
                $$lim_{Ntoinfty}S_N=1-frac{1}{2^2}=frac{3}{4}.$$






                share|cite|improve this answer









                $endgroup$
















                  7












                  7








                  7





                  $begingroup$

                  Let $$S_N=sum_{n=1}^Nleft(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}right)=frac{1}{a_1^2}-frac{1}{a_{N+1}^2}.$$
                  Since $a_1=1$ and $a_nto 2$, we have
                  $$lim_{Ntoinfty}S_N=1-frac{1}{2^2}=frac{3}{4}.$$






                  share|cite|improve this answer









                  $endgroup$



                  Let $$S_N=sum_{n=1}^Nleft(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}right)=frac{1}{a_1^2}-frac{1}{a_{N+1}^2}.$$
                  Since $a_1=1$ and $a_nto 2$, we have
                  $$lim_{Ntoinfty}S_N=1-frac{1}{2^2}=frac{3}{4}.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered yesterday









                  Eclipse SunEclipse Sun

                  7,4591437




                  7,4591437























                      4












                      $begingroup$


                      Infinite sums can behave in strange ways




                      True!




                      Sometimes more than one solution may follow logically




                      Definitely not true. There is one answer in every case (as long as you allow "the sum diverges" as an answer). Sure, there are subtle errors which may make it appear that a sum has more than one value, but that's exactly what they are - errors.



                      To avoid these errors, don't manipulate infinite sums unless you have a conclusive proof that the manipulation is justified. If you don't, go back to the definition. In your case this says
                      $$eqalign{sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)
                      &=lim_{Ntoinfty}sum_{n=1}^Nbigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)cr
                      &=lim_{Ntoinfty}biggl(frac{1}{a_1^2}-frac{1}{a_{N+1}^2}biggr) .cr}$$
                      See if you can finish the job from here.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        I stated that more than one solution may follow logically because of Grandi's series where 0, 1, and half are possible answers. Although, I suppose you could simply call the series divergent and completely avoid the controversy altogether.
                        $endgroup$
                        – s0ulr3aper07
                        yesterday






                      • 2




                        $begingroup$
                        Using standard definitions, Grandi's series is divergent and that's that. Using Cesàro summation you get $frac12$, but this does not mean the series actually converges to $frac12$. As far as I am aware, there has been absolutely no serious controversy on this question for well over a century. Well stated in Wikipedia: "the above manipulations [giving various 'sums'] do not consider what the sum of a series actually means".
                        $endgroup$
                        – David
                        yesterday


















                      4












                      $begingroup$


                      Infinite sums can behave in strange ways




                      True!




                      Sometimes more than one solution may follow logically




                      Definitely not true. There is one answer in every case (as long as you allow "the sum diverges" as an answer). Sure, there are subtle errors which may make it appear that a sum has more than one value, but that's exactly what they are - errors.



                      To avoid these errors, don't manipulate infinite sums unless you have a conclusive proof that the manipulation is justified. If you don't, go back to the definition. In your case this says
                      $$eqalign{sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)
                      &=lim_{Ntoinfty}sum_{n=1}^Nbigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)cr
                      &=lim_{Ntoinfty}biggl(frac{1}{a_1^2}-frac{1}{a_{N+1}^2}biggr) .cr}$$
                      See if you can finish the job from here.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        I stated that more than one solution may follow logically because of Grandi's series where 0, 1, and half are possible answers. Although, I suppose you could simply call the series divergent and completely avoid the controversy altogether.
                        $endgroup$
                        – s0ulr3aper07
                        yesterday






                      • 2




                        $begingroup$
                        Using standard definitions, Grandi's series is divergent and that's that. Using Cesàro summation you get $frac12$, but this does not mean the series actually converges to $frac12$. As far as I am aware, there has been absolutely no serious controversy on this question for well over a century. Well stated in Wikipedia: "the above manipulations [giving various 'sums'] do not consider what the sum of a series actually means".
                        $endgroup$
                        – David
                        yesterday
















                      4












                      4








                      4





                      $begingroup$


                      Infinite sums can behave in strange ways




                      True!




                      Sometimes more than one solution may follow logically




                      Definitely not true. There is one answer in every case (as long as you allow "the sum diverges" as an answer). Sure, there are subtle errors which may make it appear that a sum has more than one value, but that's exactly what they are - errors.



                      To avoid these errors, don't manipulate infinite sums unless you have a conclusive proof that the manipulation is justified. If you don't, go back to the definition. In your case this says
                      $$eqalign{sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)
                      &=lim_{Ntoinfty}sum_{n=1}^Nbigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)cr
                      &=lim_{Ntoinfty}biggl(frac{1}{a_1^2}-frac{1}{a_{N+1}^2}biggr) .cr}$$
                      See if you can finish the job from here.






                      share|cite|improve this answer









                      $endgroup$




                      Infinite sums can behave in strange ways




                      True!




                      Sometimes more than one solution may follow logically




                      Definitely not true. There is one answer in every case (as long as you allow "the sum diverges" as an answer). Sure, there are subtle errors which may make it appear that a sum has more than one value, but that's exactly what they are - errors.



                      To avoid these errors, don't manipulate infinite sums unless you have a conclusive proof that the manipulation is justified. If you don't, go back to the definition. In your case this says
                      $$eqalign{sum_{n=1}^inftybigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)
                      &=lim_{Ntoinfty}sum_{n=1}^Nbigg(frac{1}{a_n^2}-frac{1}{a_{n+1}^2}bigg)cr
                      &=lim_{Ntoinfty}biggl(frac{1}{a_1^2}-frac{1}{a_{N+1}^2}biggr) .cr}$$
                      See if you can finish the job from here.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered yesterday









                      DavidDavid

                      68.2k664126




                      68.2k664126












                      • $begingroup$
                        I stated that more than one solution may follow logically because of Grandi's series where 0, 1, and half are possible answers. Although, I suppose you could simply call the series divergent and completely avoid the controversy altogether.
                        $endgroup$
                        – s0ulr3aper07
                        yesterday






                      • 2




                        $begingroup$
                        Using standard definitions, Grandi's series is divergent and that's that. Using Cesàro summation you get $frac12$, but this does not mean the series actually converges to $frac12$. As far as I am aware, there has been absolutely no serious controversy on this question for well over a century. Well stated in Wikipedia: "the above manipulations [giving various 'sums'] do not consider what the sum of a series actually means".
                        $endgroup$
                        – David
                        yesterday




















                      • $begingroup$
                        I stated that more than one solution may follow logically because of Grandi's series where 0, 1, and half are possible answers. Although, I suppose you could simply call the series divergent and completely avoid the controversy altogether.
                        $endgroup$
                        – s0ulr3aper07
                        yesterday






                      • 2




                        $begingroup$
                        Using standard definitions, Grandi's series is divergent and that's that. Using Cesàro summation you get $frac12$, but this does not mean the series actually converges to $frac12$. As far as I am aware, there has been absolutely no serious controversy on this question for well over a century. Well stated in Wikipedia: "the above manipulations [giving various 'sums'] do not consider what the sum of a series actually means".
                        $endgroup$
                        – David
                        yesterday


















                      $begingroup$
                      I stated that more than one solution may follow logically because of Grandi's series where 0, 1, and half are possible answers. Although, I suppose you could simply call the series divergent and completely avoid the controversy altogether.
                      $endgroup$
                      – s0ulr3aper07
                      yesterday




                      $begingroup$
                      I stated that more than one solution may follow logically because of Grandi's series where 0, 1, and half are possible answers. Although, I suppose you could simply call the series divergent and completely avoid the controversy altogether.
                      $endgroup$
                      – s0ulr3aper07
                      yesterday




                      2




                      2




                      $begingroup$
                      Using standard definitions, Grandi's series is divergent and that's that. Using Cesàro summation you get $frac12$, but this does not mean the series actually converges to $frac12$. As far as I am aware, there has been absolutely no serious controversy on this question for well over a century. Well stated in Wikipedia: "the above manipulations [giving various 'sums'] do not consider what the sum of a series actually means".
                      $endgroup$
                      – David
                      yesterday






                      $begingroup$
                      Using standard definitions, Grandi's series is divergent and that's that. Using Cesàro summation you get $frac12$, but this does not mean the series actually converges to $frac12$. As far as I am aware, there has been absolutely no serious controversy on this question for well over a century. Well stated in Wikipedia: "the above manipulations [giving various 'sums'] do not consider what the sum of a series actually means".
                      $endgroup$
                      – David
                      yesterday













                      3












                      $begingroup$

                      $a_n$ does not approach zero,
                      so the end term can not
                      be disregarded.



                      In other words,
                      the sum up to $n$ is
                      $frac1{a_1^2}
                      -frac1{a_n^2}
                      to 1-frac14
                      =frac34
                      $                      .






                      share|cite|improve this answer









                      $endgroup$


















                        3












                        $begingroup$

                        $a_n$ does not approach zero,
                        so the end term can not
                        be disregarded.



                        In other words,
                        the sum up to $n$ is
                        $frac1{a_1^2}
                        -frac1{a_n^2}
                        to 1-frac14
                        =frac34
                        $                        .






                        share|cite|improve this answer









                        $endgroup$
















                          3












                          3








                          3





                          $begingroup$

                          $a_n$ does not approach zero,
                          so the end term can not
                          be disregarded.



                          In other words,
                          the sum up to $n$ is
                          $frac1{a_1^2}
                          -frac1{a_n^2}
                          to 1-frac14
                          =frac34
                          $                          .






                          share|cite|improve this answer









                          $endgroup$



                          $a_n$ does not approach zero,
                          so the end term can not
                          be disregarded.



                          In other words,
                          the sum up to $n$ is
                          $frac1{a_1^2}
                          -frac1{a_n^2}
                          to 1-frac14
                          =frac34
                          $                          .







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered yesterday









                          marty cohenmarty cohen

                          73.4k549128




                          73.4k549128






























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                              RAC Tourist Trophy

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                              Pour les articles homonymes, voir Tourist Trophy. Le RAC Tourist Trophy est une course automobile organisée par le Royal Automobile Club d'Angleterre. Créée en 1905, cette course a fait partie des épreuves comptant pour le championnat du monde des voitures de sport entre 1953 et 1955, en 1958 et 1959, et entre 1962 et 1965. Après une interruption de 17 ans (l'épreuve s'arrête après 1988), le RAC Tourist Trophy reprend en 2005 sous l'égide de la FIA GT. Palmarès | John Napier sur Arrol-Johnston, vainqueur du premier RAC Tourist Trophy, en 1905. L'honorable C.S. Rolls, vainqueur du RAC Tourist Trophy 1906. Ernest Curtis, sur Rover 20 hp, vainqueur du RAC Tourist Trophy 1907. William Watson, vainqueur du RAC Tourist Trophy 1908. Kenelm Lee Guinness, vainqueur du RAC Tourist Trophy 1914 sur Sunbeam. Départ du RAC Tourist Trophy 1928. Louis Gérard, vainqueur du RAC Tourist Trophy 1938 sur Delage. Année Circuit Vainqueur
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