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







share|cite|improve this question









$endgroup$

















    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













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 1 hour ago









      s0ulr3aper07s0ulr3aper07

      3069




      3069






















          3 Answers
          3






          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









          $endgroup$





















            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
                29 mins ago











              Your Answer





              StackExchange.ifUsing("editor", function () {
              return StackExchange.using("mathjaxEditing", function () {
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              });
              });
              }, "mathjax-editing");

              StackExchange.ready(function() {
              var channelOptions = {
              tags: "".split(" "),
              id: "69"
              };
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function() {
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled) {
              StackExchange.using("snippets", function() {
              createEditor();
              });
              }
              else {
              createEditor();
              }
              });

              function createEditor() {
              StackExchange.prepareEditor({
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader: {
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              },
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              });


              }
              });














              draft saved

              draft discarded


















              StackExchange.ready(
              function () {
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3099389%2fconvergence-of-this-particular-series%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              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 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
                                29 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
                                29 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
                                29 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
                                29 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
                              29 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
                              29 mins ago


















                              draft saved

                              draft discarded




















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid …



                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.


                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3099389%2fconvergence-of-this-particular-series%23new-answer', 'question_page');
                              }
                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              -

                              Popular posts from this blog

                              Bundesstraße 106

                              Image
                              Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
                              Read more

                              Verónica Boquete

                              Image
                              Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
                              Read more

                              Ida-Boy-Ed-Garten

                              Image
                              Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
                              Read more