Evaluate $sum_{n=1}^{infty} frac{2n+1}{(n^{2} +n)^{2}}$












1












$begingroup$



Evaluate $$sum_{n=1}^{infty} frac{2n+1}{(n^{2}+n)^{2}}.$$




I am getting two different results by using two different methods -



First Method



The above sum can be written as



begin{align}sum_{n=1}^{N} (1/n^{2} - 1/(n+1)^{2})&= 1 - 1/4 + 1/4 - 1/9 dots -1/(N+1)^{2}\ &= 1 - 1/(N+1)^{2} end{align}



Taking the limit as $Ntoinfty$, we have the the sum equal to $1$.



Second Method



Above sum is equal to -
begin{align}int_{1}^{infty} frac{2x+1}{(x^{2}+x)^{2}},dxend{align}



Put $x^2 + x = t$



$$int_{2}^{infty} dt/t^{2}$$



$=[-1/t]_{2}^{infty}$



$=1/2$



Why are these methods are giving different results?






real-analysis







share|cite|improve this question











$endgroup$








  • 10




    $begingroup$
    The sum and the integral are not equal. That is why you're getting different answers.
    $endgroup$
    – Clayton
    Dec 24 '18 at 17:32






  • 2




    $begingroup$
    For n=1, the first term would be $3/4$. You missed something in your second method.
    $endgroup$
    – Love Invariants
    Dec 24 '18 at 17:34










  • $begingroup$
    @Clayton By cauchy integral test, this sum and the integral (in second method) should behave alike. So, why there values are different?
    $endgroup$
    – Mathsaddict
    Dec 24 '18 at 17:40






  • 3




    $begingroup$
    They behave similarly, but that doesn't imply they represent the same numbers.
    $endgroup$
    – Clayton
    Dec 24 '18 at 17:41










  • $begingroup$
    A visual argument might help to see why the two are fundamentally different.
    $endgroup$
    – SvanN
    Dec 24 '18 at 17:44
















1












$begingroup$



Evaluate $$sum_{n=1}^{infty} frac{2n+1}{(n^{2}+n)^{2}}.$$




I am getting two different results by using two different methods -



First Method



The above sum can be written as



begin{align}sum_{n=1}^{N} (1/n^{2} - 1/(n+1)^{2})&= 1 - 1/4 + 1/4 - 1/9 dots -1/(N+1)^{2}\ &= 1 - 1/(N+1)^{2} end{align}



Taking the limit as $Ntoinfty$, we have the the sum equal to $1$.



Second Method



Above sum is equal to -
begin{align}int_{1}^{infty} frac{2x+1}{(x^{2}+x)^{2}},dxend{align}



Put $x^2 + x = t$



$$int_{2}^{infty} dt/t^{2}$$



$=[-1/t]_{2}^{infty}$



$=1/2$



Why are these methods are giving different results?






real-analysis







share|cite|improve this question











$endgroup$








  • 10




    $begingroup$
    The sum and the integral are not equal. That is why you're getting different answers.
    $endgroup$
    – Clayton
    Dec 24 '18 at 17:32






  • 2




    $begingroup$
    For n=1, the first term would be $3/4$. You missed something in your second method.
    $endgroup$
    – Love Invariants
    Dec 24 '18 at 17:34










  • $begingroup$
    @Clayton By cauchy integral test, this sum and the integral (in second method) should behave alike. So, why there values are different?
    $endgroup$
    – Mathsaddict
    Dec 24 '18 at 17:40






  • 3




    $begingroup$
    They behave similarly, but that doesn't imply they represent the same numbers.
    $endgroup$
    – Clayton
    Dec 24 '18 at 17:41










  • $begingroup$
    A visual argument might help to see why the two are fundamentally different.
    $endgroup$
    – SvanN
    Dec 24 '18 at 17:44














1












1








1


1



$begingroup$



Evaluate $$sum_{n=1}^{infty} frac{2n+1}{(n^{2}+n)^{2}}.$$




I am getting two different results by using two different methods -



First Method



The above sum can be written as



begin{align}sum_{n=1}^{N} (1/n^{2} - 1/(n+1)^{2})&= 1 - 1/4 + 1/4 - 1/9 dots -1/(N+1)^{2}\ &= 1 - 1/(N+1)^{2} end{align}



Taking the limit as $Ntoinfty$, we have the the sum equal to $1$.



Second Method



Above sum is equal to -
begin{align}int_{1}^{infty} frac{2x+1}{(x^{2}+x)^{2}},dxend{align}



Put $x^2 + x = t$



$$int_{2}^{infty} dt/t^{2}$$



$=[-1/t]_{2}^{infty}$



$=1/2$



Why are these methods are giving different results?






real-analysis







share|cite|improve this question











$endgroup$





Evaluate $$sum_{n=1}^{infty} frac{2n+1}{(n^{2}+n)^{2}}.$$




I am getting two different results by using two different methods -



First Method



The above sum can be written as



begin{align}sum_{n=1}^{N} (1/n^{2} - 1/(n+1)^{2})&= 1 - 1/4 + 1/4 - 1/9 dots -1/(N+1)^{2}\ &= 1 - 1/(N+1)^{2} end{align}



Taking the limit as $Ntoinfty$, we have the the sum equal to $1$.



Second Method



Above sum is equal to -
begin{align}int_{1}^{infty} frac{2x+1}{(x^{2}+x)^{2}},dxend{align}



Put $x^2 + x = t$



$$int_{2}^{infty} dt/t^{2}$$



$=[-1/t]_{2}^{infty}$



$=1/2$



Why are these methods are giving different results?






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 24 '18 at 17:38









Clayton

19.3k33287




19.3k33287










asked Dec 24 '18 at 17:31









MathsaddictMathsaddict

3669




3669








  • 10




    $begingroup$
    The sum and the integral are not equal. That is why you're getting different answers.
    $endgroup$
    – Clayton
    Dec 24 '18 at 17:32






  • 2




    $begingroup$
    For n=1, the first term would be $3/4$. You missed something in your second method.
    $endgroup$
    – Love Invariants
    Dec 24 '18 at 17:34










  • $begingroup$
    @Clayton By cauchy integral test, this sum and the integral (in second method) should behave alike. So, why there values are different?
    $endgroup$
    – Mathsaddict
    Dec 24 '18 at 17:40






  • 3




    $begingroup$
    They behave similarly, but that doesn't imply they represent the same numbers.
    $endgroup$
    – Clayton
    Dec 24 '18 at 17:41










  • $begingroup$
    A visual argument might help to see why the two are fundamentally different.
    $endgroup$
    – SvanN
    Dec 24 '18 at 17:44














  • 10




    $begingroup$
    The sum and the integral are not equal. That is why you're getting different answers.
    $endgroup$
    – Clayton
    Dec 24 '18 at 17:32






  • 2




    $begingroup$
    For n=1, the first term would be $3/4$. You missed something in your second method.
    $endgroup$
    – Love Invariants
    Dec 24 '18 at 17:34










  • $begingroup$
    @Clayton By cauchy integral test, this sum and the integral (in second method) should behave alike. So, why there values are different?
    $endgroup$
    – Mathsaddict
    Dec 24 '18 at 17:40






  • 3




    $begingroup$
    They behave similarly, but that doesn't imply they represent the same numbers.
    $endgroup$
    – Clayton
    Dec 24 '18 at 17:41










  • $begingroup$
    A visual argument might help to see why the two are fundamentally different.
    $endgroup$
    – SvanN
    Dec 24 '18 at 17:44








10




10




$begingroup$
The sum and the integral are not equal. That is why you're getting different answers.
$endgroup$
– Clayton
Dec 24 '18 at 17:32




$begingroup$
The sum and the integral are not equal. That is why you're getting different answers.
$endgroup$
– Clayton
Dec 24 '18 at 17:32




2




2




$begingroup$
For n=1, the first term would be $3/4$. You missed something in your second method.
$endgroup$
– Love Invariants
Dec 24 '18 at 17:34




$begingroup$
For n=1, the first term would be $3/4$. You missed something in your second method.
$endgroup$
– Love Invariants
Dec 24 '18 at 17:34












$begingroup$
@Clayton By cauchy integral test, this sum and the integral (in second method) should behave alike. So, why there values are different?
$endgroup$
– Mathsaddict
Dec 24 '18 at 17:40




$begingroup$
@Clayton By cauchy integral test, this sum and the integral (in second method) should behave alike. So, why there values are different?
$endgroup$
– Mathsaddict
Dec 24 '18 at 17:40




3




3




$begingroup$
They behave similarly, but that doesn't imply they represent the same numbers.
$endgroup$
– Clayton
Dec 24 '18 at 17:41




$begingroup$
They behave similarly, but that doesn't imply they represent the same numbers.
$endgroup$
– Clayton
Dec 24 '18 at 17:41












$begingroup$
A visual argument might help to see why the two are fundamentally different.
$endgroup$
– SvanN
Dec 24 '18 at 17:44




$begingroup$
A visual argument might help to see why the two are fundamentally different.
$endgroup$
– SvanN
Dec 24 '18 at 17:44










2 Answers
2






active

oldest

votes


















3












$begingroup$

The Integral Test for series does NOT assert that the infinite series and the improper integral are equal to each other. In fact, from the proof, it can be seen that they can't possibly be equal to each other (except for maybe some carefully constructed step functions, if we relax the continuity requirement). The test only says that if the integral converges, then the series converges as well (and the same for divergence), but it does not provide a value for the series in this case (although it does provide a useful remainder estimate). In short:
$$sum_{n=1}^{infty}f(n)color{red}{neq}int_1^{infty}f(x),dx,$$
which is why your second "method" is wrong.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    In context, hopefully not too trivial.



    Let $f(n)=dfrac{2n+1}{(n^2+n)^2}$, $f(n)$ is strictly decreasing.



    1)Your sum $sum_{1}^{infty}f(n)$ is an upper sum



    for the integral $int_{1}^{infty}f(x)dx$.



    $U :=sum_{1}^{infty}f(n)=1$;



    2)Now consider the lower sum:



    $sum_{2}^{infty}f(n)$ for the integral.



    $L := sum_{2}^{infty}f(n)=1/4;$



    We have



    $L =1/4 < 1/2$ (Integral)$ < U=1$.



    See:
    https://en.m.wikipedia.org/wiki/Integral_test_for_convergence



    Link given by Zipirovic.






    share|cite|improve this answer









    $endgroup$













      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%2f3051479%2fevaluate-sum-n-1-infty-frac2n1n2-n2%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      The Integral Test for series does NOT assert that the infinite series and the improper integral are equal to each other. In fact, from the proof, it can be seen that they can't possibly be equal to each other (except for maybe some carefully constructed step functions, if we relax the continuity requirement). The test only says that if the integral converges, then the series converges as well (and the same for divergence), but it does not provide a value for the series in this case (although it does provide a useful remainder estimate). In short:
      $$sum_{n=1}^{infty}f(n)color{red}{neq}int_1^{infty}f(x),dx,$$
      which is why your second "method" is wrong.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        The Integral Test for series does NOT assert that the infinite series and the improper integral are equal to each other. In fact, from the proof, it can be seen that they can't possibly be equal to each other (except for maybe some carefully constructed step functions, if we relax the continuity requirement). The test only says that if the integral converges, then the series converges as well (and the same for divergence), but it does not provide a value for the series in this case (although it does provide a useful remainder estimate). In short:
        $$sum_{n=1}^{infty}f(n)color{red}{neq}int_1^{infty}f(x),dx,$$
        which is why your second "method" is wrong.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          The Integral Test for series does NOT assert that the infinite series and the improper integral are equal to each other. In fact, from the proof, it can be seen that they can't possibly be equal to each other (except for maybe some carefully constructed step functions, if we relax the continuity requirement). The test only says that if the integral converges, then the series converges as well (and the same for divergence), but it does not provide a value for the series in this case (although it does provide a useful remainder estimate). In short:
          $$sum_{n=1}^{infty}f(n)color{red}{neq}int_1^{infty}f(x),dx,$$
          which is why your second "method" is wrong.






          share|cite|improve this answer









          $endgroup$



          The Integral Test for series does NOT assert that the infinite series and the improper integral are equal to each other. In fact, from the proof, it can be seen that they can't possibly be equal to each other (except for maybe some carefully constructed step functions, if we relax the continuity requirement). The test only says that if the integral converges, then the series converges as well (and the same for divergence), but it does not provide a value for the series in this case (although it does provide a useful remainder estimate). In short:
          $$sum_{n=1}^{infty}f(n)color{red}{neq}int_1^{infty}f(x),dx,$$
          which is why your second "method" is wrong.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 24 '18 at 17:43









          zipirovichzipirovich

          11.3k11731




          11.3k11731























              0












              $begingroup$

              In context, hopefully not too trivial.



              Let $f(n)=dfrac{2n+1}{(n^2+n)^2}$, $f(n)$ is strictly decreasing.



              1)Your sum $sum_{1}^{infty}f(n)$ is an upper sum



              for the integral $int_{1}^{infty}f(x)dx$.



              $U :=sum_{1}^{infty}f(n)=1$;



              2)Now consider the lower sum:



              $sum_{2}^{infty}f(n)$ for the integral.



              $L := sum_{2}^{infty}f(n)=1/4;$



              We have



              $L =1/4 < 1/2$ (Integral)$ < U=1$.



              See:
              https://en.m.wikipedia.org/wiki/Integral_test_for_convergence



              Link given by Zipirovic.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                In context, hopefully not too trivial.



                Let $f(n)=dfrac{2n+1}{(n^2+n)^2}$, $f(n)$ is strictly decreasing.



                1)Your sum $sum_{1}^{infty}f(n)$ is an upper sum



                for the integral $int_{1}^{infty}f(x)dx$.



                $U :=sum_{1}^{infty}f(n)=1$;



                2)Now consider the lower sum:



                $sum_{2}^{infty}f(n)$ for the integral.



                $L := sum_{2}^{infty}f(n)=1/4;$



                We have



                $L =1/4 < 1/2$ (Integral)$ < U=1$.



                See:
                https://en.m.wikipedia.org/wiki/Integral_test_for_convergence



                Link given by Zipirovic.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  In context, hopefully not too trivial.



                  Let $f(n)=dfrac{2n+1}{(n^2+n)^2}$, $f(n)$ is strictly decreasing.



                  1)Your sum $sum_{1}^{infty}f(n)$ is an upper sum



                  for the integral $int_{1}^{infty}f(x)dx$.



                  $U :=sum_{1}^{infty}f(n)=1$;



                  2)Now consider the lower sum:



                  $sum_{2}^{infty}f(n)$ for the integral.



                  $L := sum_{2}^{infty}f(n)=1/4;$



                  We have



                  $L =1/4 < 1/2$ (Integral)$ < U=1$.



                  See:
                  https://en.m.wikipedia.org/wiki/Integral_test_for_convergence



                  Link given by Zipirovic.






                  share|cite|improve this answer









                  $endgroup$



                  In context, hopefully not too trivial.



                  Let $f(n)=dfrac{2n+1}{(n^2+n)^2}$, $f(n)$ is strictly decreasing.



                  1)Your sum $sum_{1}^{infty}f(n)$ is an upper sum



                  for the integral $int_{1}^{infty}f(x)dx$.



                  $U :=sum_{1}^{infty}f(n)=1$;



                  2)Now consider the lower sum:



                  $sum_{2}^{infty}f(n)$ for the integral.



                  $L := sum_{2}^{infty}f(n)=1/4;$



                  We have



                  $L =1/4 < 1/2$ (Integral)$ < U=1$.



                  See:
                  https://en.m.wikipedia.org/wiki/Integral_test_for_convergence



                  Link given by Zipirovic.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 24 '18 at 18:34









                  Peter SzilasPeter Szilas

                  11.5k2822




                  11.5k2822






























                      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%2f3051479%2fevaluate-sum-n-1-infty-frac2n1n2-n2%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

                      Probability when a professor distributes a quiz and homework assignment to a class of n students.

                      Image
                      0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
                      Read more

                      Aardman Animations

                      Image
                      Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
                      Read more

                      Are they similar matrix

                      Image
                      0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
                      Read more