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









    $endgroup$













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      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






























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