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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    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 1 hour ago









      s0ulr3aper07s0ulr3aper07

      3069




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          3 Answers
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          $begingroup$

          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









              $endgroup$













              • $begingroup$
                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
                28 mins ago











              Your Answer





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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

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              active

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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









              $endgroup$


















                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 58 mins ago









                  Eclipse SunEclipse Sun

                  7,3841437




                  7,3841437























                      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 55 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









                              $endgroup$













                              • $begingroup$
                                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
                                28 mins ago
















                              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









                              $endgroup$













                              • $begingroup$
                                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
                                28 mins ago














                              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 55 mins ago









                              DavidDavid

                              68.1k664126




                              68.1k664126












                              • $begingroup$
                                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
                                28 mins ago


















                              • $begingroup$
                                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
                                28 mins ago
















                              $begingroup$
                              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
                              28 mins ago




                              $begingroup$
                              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
                              28 mins ago


















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                              濃尾地震

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                              濃尾地震 『岐阜市街大地震之図』 歌川国利画 本震 発生日 1891年（明治24年）10月28日 発生時刻 6時38分50秒（JST） 震央 日本 岐阜県本巣郡西根尾村（現・本巣市） 北緯35度35分 東経136度20分（ 地図 ） 規模    マグニチュード（M）8.0 最大震度    震度7： 注1 福井県今立郡鯖江町、愛知県葉栗郡大田島村、東春日井郡勝川村 地震の種類 直下型地震 被害 死傷者数 死者7,273人、負傷者17,175人 注1：当時の震度階級では「激烈」 プロジェクト:地球科学 プロジェクト:災害 テンプレートを表示 濃尾地震 （のうびじしん）は、1891年（明治24年）10月28日に濃尾地方で発生した、日本史上最大の内陸地殻内地震。「 美濃・尾張地震 （みの・おわりじしん）」とも呼ばれている。辛卯の年に発生したことから 辛卯震災 と呼んでいる報告書もある。 目次 1 概要 1.1 震源断層 2 被害 3 各地の震度 4 前兆現象 5 報道 6 学術的な意義 7 地震防災 8 脚注 9 関連項目 10 外部リンク 概要 濃尾地震発生当時の根尾谷断層 濃尾地震を引き起こした根尾谷断層 写真中央を斜めに走る段差が根尾谷断層 濃尾地震は、1891年10月28日6時38分50秒に発生した。震源は、岐阜県本巣郡西根尾村（現・本巣市）、北緯35度35分、東経136度20分付近。河角廣 (1951) は岐阜市付近（北緯35.6度、東経136.6度）に震央を仮定し規模 M K = 7.0 を与え [1] 、マグニチュードは M = 8.4 に換算されているが、明治・大正期の地震については0.5程度大きく見積もられているとされる [2] 。また、震央距離と震度との関係など当時のデータから後にM8.0 [3] とも推定される。アメリカ地質調査所 (USGS) でも8.0としている [4] 。「根尾谷断層帯」が活動した典型的な内陸地殻内地震（いわゆる直下型地震）であり、これは記録が残っている日本の内陸
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