Convergence of this particular series












1












$begingroup$


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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$endgroup$

















    1












    $begingroup$


    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









    $endgroup$















      1












      1








      1





      $begingroup$


      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













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      asked 35 mins ago









      s0ulr3aper07s0ulr3aper07

      3049




      3049






















          3 Answers
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          2












          $begingroup$

          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









          $endgroup$





















            1












            $begingroup$

            $a_n$ 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









            $endgroup$





















              1












              $begingroup$


              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









              $endgroup$













                Your Answer





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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                2












                $begingroup$

                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









                $endgroup$


















                  2












                  $begingroup$

                  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









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    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 29 mins ago









                    Eclipse SunEclipse Sun

                    7,3691437




                    7,3691437























                        1












                        $begingroup$

                        $a_n$ 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









                        $endgroup$


















                          1












                          $begingroup$

                          $a_n$ 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









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            $a_n$ 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 26 mins ago









                            marty cohenmarty cohen

                            73.4k549128




                            73.4k549128























                                1












                                $begingroup$


                                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









                                $endgroup$


















                                  1












                                  $begingroup$


                                  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









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$


                                    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 25 mins ago









                                    DavidDavid

                                    68.1k664126




                                    68.1k664126






























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