Why is the inequality $sum_{n=1}^{infty} frac{1}{n^2} leq 1 + int_1^{infty} frac{1}{x^2}$ true?












2














$$sum_{n=1}^{infty} frac{1}{n^2} leq 1 + int_1^{infty} frac{1}{x^2}dx$$



I'm having trouble figuring out why the inequality above is true. I understand the following inequality:



$$int_1^{infty} frac{1}{x^2}dx leq sum_{n=1}^{infty} frac{1}{n^2}$$



It makes sense because the rectangles formed from the right side of the inequality have pieces that go over-top of the function like so:



enter image description here



So let's say I rewrite $sum_{n=1}^{infty} frac{1}{n^2}$ as $1 + sum_{n=2}^{infty} frac{1}{n^2}$ since they are equivalent.



Why is that less than $1 + int_1^{infty} frac{1}{x^2}dx$?



If I picture it in my head it sounds like its saying that the first term is greater than all the little pieces that form above the curve in the picture above.



My claims come specifically from page 60 of this webpage from Dartmouth






integration sequences-and-series inequality summation power-series







share|cite|improve this question






















  • Draw similar but shorter rectangles.
    – Lord Shark the Unknown
    Nov 27 '18 at 18:47
















2














$$sum_{n=1}^{infty} frac{1}{n^2} leq 1 + int_1^{infty} frac{1}{x^2}dx$$



I'm having trouble figuring out why the inequality above is true. I understand the following inequality:



$$int_1^{infty} frac{1}{x^2}dx leq sum_{n=1}^{infty} frac{1}{n^2}$$



It makes sense because the rectangles formed from the right side of the inequality have pieces that go over-top of the function like so:



enter image description here



So let's say I rewrite $sum_{n=1}^{infty} frac{1}{n^2}$ as $1 + sum_{n=2}^{infty} frac{1}{n^2}$ since they are equivalent.



Why is that less than $1 + int_1^{infty} frac{1}{x^2}dx$?



If I picture it in my head it sounds like its saying that the first term is greater than all the little pieces that form above the curve in the picture above.



My claims come specifically from page 60 of this webpage from Dartmouth






integration sequences-and-series inequality summation power-series







share|cite|improve this question






















  • Draw similar but shorter rectangles.
    – Lord Shark the Unknown
    Nov 27 '18 at 18:47














2












2








2







$$sum_{n=1}^{infty} frac{1}{n^2} leq 1 + int_1^{infty} frac{1}{x^2}dx$$



I'm having trouble figuring out why the inequality above is true. I understand the following inequality:



$$int_1^{infty} frac{1}{x^2}dx leq sum_{n=1}^{infty} frac{1}{n^2}$$



It makes sense because the rectangles formed from the right side of the inequality have pieces that go over-top of the function like so:



enter image description here



So let's say I rewrite $sum_{n=1}^{infty} frac{1}{n^2}$ as $1 + sum_{n=2}^{infty} frac{1}{n^2}$ since they are equivalent.



Why is that less than $1 + int_1^{infty} frac{1}{x^2}dx$?



If I picture it in my head it sounds like its saying that the first term is greater than all the little pieces that form above the curve in the picture above.



My claims come specifically from page 60 of this webpage from Dartmouth






integration sequences-and-series inequality summation power-series







share|cite|improve this question













$$sum_{n=1}^{infty} frac{1}{n^2} leq 1 + int_1^{infty} frac{1}{x^2}dx$$



I'm having trouble figuring out why the inequality above is true. I understand the following inequality:



$$int_1^{infty} frac{1}{x^2}dx leq sum_{n=1}^{infty} frac{1}{n^2}$$



It makes sense because the rectangles formed from the right side of the inequality have pieces that go over-top of the function like so:



enter image description here



So let's say I rewrite $sum_{n=1}^{infty} frac{1}{n^2}$ as $1 + sum_{n=2}^{infty} frac{1}{n^2}$ since they are equivalent.



Why is that less than $1 + int_1^{infty} frac{1}{x^2}dx$?



If I picture it in my head it sounds like its saying that the first term is greater than all the little pieces that form above the curve in the picture above.



My claims come specifically from page 60 of this webpage from Dartmouth






integration sequences-and-series inequality summation power-series




integration sequences-and-series inequality summation power-series






share|cite|improve this question













share|cite|improve this question











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asked Nov 27 '18 at 18:44









James Mitchell

26827




26827












  • Draw similar but shorter rectangles.
    – Lord Shark the Unknown
    Nov 27 '18 at 18:47


















  • Draw similar but shorter rectangles.
    – Lord Shark the Unknown
    Nov 27 '18 at 18:47
















Draw similar but shorter rectangles.
– Lord Shark the Unknown
Nov 27 '18 at 18:47




Draw similar but shorter rectangles.
– Lord Shark the Unknown
Nov 27 '18 at 18:47










4 Answers
4






active

oldest

votes


















2














The right endpoint sums for the integral have the form:



$$
sum_{n = 2}^infty frac{1}{n^2}
$$
and we know:
$$
sum_{n = 2}^infty frac{1}{n^2} leq int_1^infty frac{1}{x^2} leq sum_{n = 1}^infty frac{1}{n^2}
$$
Subtracting the RHS, we have:
$$
-1 leq int_1^infty frac{1}{x^2} - sum_{n = 1}^infty
frac{1}{n^2}$$
Multiplying by $-1$, we have:



$$
1 geq sum_{n = 1}^infty
frac{1}{n^2} -int_1^infty frac{1}{x^2}
$$
as we wanted.






share|cite|improve this answer





























    1














    A way to see that from the graph is as follows



    $$sum_{n=1}^{infty} frac{1}{n^2} leq 1 + int_2^{infty} overbrace{frac{1}{(x-1)^2}}^{graph, for, frac1{x^2},shifted , by, 1}dx= 1 + int_1^{infty} frac{1}{x^2}dx$$






    share|cite|improve this answer





























      1














      Hint: Subract $1$ from both sides to see inequality is the same as



      $$sum_{n=2}^{infty} frac{1}{n^2} le int_1^inftyfrac{dx}{x^2}.$$



      Now do your rectangle comparisons.






      share|cite|improve this answer





























        0














        More generally, suppose $f$ is strictly decreasing on $xge 1$, so any positive integer $n$ satisfies $f(n+1)leint_n^{n+1}f(x)dxle f(n)$. Summing, $sum_{nge 2}f(n)leint_1^infty f(x)dxlesum_{nge 1}f(n)$. Equivalently, $int_1^infty f(x)dxlesum_{nge 1}f(n)le f(1)+int_1^infty f(x)dx$. You just need the choice $f(x)=x^{-2}$. Another important corollary, called the integral test, is that for such $f$ the series $sum_{nge 1}f(n)$ converges iff $int_1^infty f(x)dx$ does. In particular, the divergence of the harmonic series is equivalent to $int_1^inftyfrac{dx}{x}=lninfty=infty$.






        share|cite|improve this answer





















          Your Answer





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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2














          The right endpoint sums for the integral have the form:



          $$
          sum_{n = 2}^infty frac{1}{n^2}
          $$
          and we know:
          $$
          sum_{n = 2}^infty frac{1}{n^2} leq int_1^infty frac{1}{x^2} leq sum_{n = 1}^infty frac{1}{n^2}
          $$
          Subtracting the RHS, we have:
          $$
          -1 leq int_1^infty frac{1}{x^2} - sum_{n = 1}^infty
          frac{1}{n^2}$$
          Multiplying by $-1$, we have:



          $$
          1 geq sum_{n = 1}^infty
          frac{1}{n^2} -int_1^infty frac{1}{x^2}
          $$
          as we wanted.






          share|cite|improve this answer


























            2














            The right endpoint sums for the integral have the form:



            $$
            sum_{n = 2}^infty frac{1}{n^2}
            $$
            and we know:
            $$
            sum_{n = 2}^infty frac{1}{n^2} leq int_1^infty frac{1}{x^2} leq sum_{n = 1}^infty frac{1}{n^2}
            $$
            Subtracting the RHS, we have:
            $$
            -1 leq int_1^infty frac{1}{x^2} - sum_{n = 1}^infty
            frac{1}{n^2}$$
            Multiplying by $-1$, we have:



            $$
            1 geq sum_{n = 1}^infty
            frac{1}{n^2} -int_1^infty frac{1}{x^2}
            $$
            as we wanted.






            share|cite|improve this answer
























              2












              2








              2






              The right endpoint sums for the integral have the form:



              $$
              sum_{n = 2}^infty frac{1}{n^2}
              $$
              and we know:
              $$
              sum_{n = 2}^infty frac{1}{n^2} leq int_1^infty frac{1}{x^2} leq sum_{n = 1}^infty frac{1}{n^2}
              $$
              Subtracting the RHS, we have:
              $$
              -1 leq int_1^infty frac{1}{x^2} - sum_{n = 1}^infty
              frac{1}{n^2}$$
              Multiplying by $-1$, we have:



              $$
              1 geq sum_{n = 1}^infty
              frac{1}{n^2} -int_1^infty frac{1}{x^2}
              $$
              as we wanted.






              share|cite|improve this answer












              The right endpoint sums for the integral have the form:



              $$
              sum_{n = 2}^infty frac{1}{n^2}
              $$
              and we know:
              $$
              sum_{n = 2}^infty frac{1}{n^2} leq int_1^infty frac{1}{x^2} leq sum_{n = 1}^infty frac{1}{n^2}
              $$
              Subtracting the RHS, we have:
              $$
              -1 leq int_1^infty frac{1}{x^2} - sum_{n = 1}^infty
              frac{1}{n^2}$$
              Multiplying by $-1$, we have:



              $$
              1 geq sum_{n = 1}^infty
              frac{1}{n^2} -int_1^infty frac{1}{x^2}
              $$
              as we wanted.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 27 '18 at 19:01









              rubikscube09

              1,169717




              1,169717























                  1














                  A way to see that from the graph is as follows



                  $$sum_{n=1}^{infty} frac{1}{n^2} leq 1 + int_2^{infty} overbrace{frac{1}{(x-1)^2}}^{graph, for, frac1{x^2},shifted , by, 1}dx= 1 + int_1^{infty} frac{1}{x^2}dx$$






                  share|cite|improve this answer


























                    1














                    A way to see that from the graph is as follows



                    $$sum_{n=1}^{infty} frac{1}{n^2} leq 1 + int_2^{infty} overbrace{frac{1}{(x-1)^2}}^{graph, for, frac1{x^2},shifted , by, 1}dx= 1 + int_1^{infty} frac{1}{x^2}dx$$






                    share|cite|improve this answer
























                      1












                      1








                      1






                      A way to see that from the graph is as follows



                      $$sum_{n=1}^{infty} frac{1}{n^2} leq 1 + int_2^{infty} overbrace{frac{1}{(x-1)^2}}^{graph, for, frac1{x^2},shifted , by, 1}dx= 1 + int_1^{infty} frac{1}{x^2}dx$$






                      share|cite|improve this answer












                      A way to see that from the graph is as follows



                      $$sum_{n=1}^{infty} frac{1}{n^2} leq 1 + int_2^{infty} overbrace{frac{1}{(x-1)^2}}^{graph, for, frac1{x^2},shifted , by, 1}dx= 1 + int_1^{infty} frac{1}{x^2}dx$$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 27 '18 at 18:47









                      gimusi

                      1




                      1























                          1














                          Hint: Subract $1$ from both sides to see inequality is the same as



                          $$sum_{n=2}^{infty} frac{1}{n^2} le int_1^inftyfrac{dx}{x^2}.$$



                          Now do your rectangle comparisons.






                          share|cite|improve this answer


























                            1














                            Hint: Subract $1$ from both sides to see inequality is the same as



                            $$sum_{n=2}^{infty} frac{1}{n^2} le int_1^inftyfrac{dx}{x^2}.$$



                            Now do your rectangle comparisons.






                            share|cite|improve this answer
























                              1












                              1








                              1






                              Hint: Subract $1$ from both sides to see inequality is the same as



                              $$sum_{n=2}^{infty} frac{1}{n^2} le int_1^inftyfrac{dx}{x^2}.$$



                              Now do your rectangle comparisons.






                              share|cite|improve this answer












                              Hint: Subract $1$ from both sides to see inequality is the same as



                              $$sum_{n=2}^{infty} frac{1}{n^2} le int_1^inftyfrac{dx}{x^2}.$$



                              Now do your rectangle comparisons.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Nov 27 '18 at 19:09









                              zhw.

                              71.6k43075




                              71.6k43075























                                  0














                                  More generally, suppose $f$ is strictly decreasing on $xge 1$, so any positive integer $n$ satisfies $f(n+1)leint_n^{n+1}f(x)dxle f(n)$. Summing, $sum_{nge 2}f(n)leint_1^infty f(x)dxlesum_{nge 1}f(n)$. Equivalently, $int_1^infty f(x)dxlesum_{nge 1}f(n)le f(1)+int_1^infty f(x)dx$. You just need the choice $f(x)=x^{-2}$. Another important corollary, called the integral test, is that for such $f$ the series $sum_{nge 1}f(n)$ converges iff $int_1^infty f(x)dx$ does. In particular, the divergence of the harmonic series is equivalent to $int_1^inftyfrac{dx}{x}=lninfty=infty$.






                                  share|cite|improve this answer


























                                    0














                                    More generally, suppose $f$ is strictly decreasing on $xge 1$, so any positive integer $n$ satisfies $f(n+1)leint_n^{n+1}f(x)dxle f(n)$. Summing, $sum_{nge 2}f(n)leint_1^infty f(x)dxlesum_{nge 1}f(n)$. Equivalently, $int_1^infty f(x)dxlesum_{nge 1}f(n)le f(1)+int_1^infty f(x)dx$. You just need the choice $f(x)=x^{-2}$. Another important corollary, called the integral test, is that for such $f$ the series $sum_{nge 1}f(n)$ converges iff $int_1^infty f(x)dx$ does. In particular, the divergence of the harmonic series is equivalent to $int_1^inftyfrac{dx}{x}=lninfty=infty$.






                                    share|cite|improve this answer
























                                      0












                                      0








                                      0






                                      More generally, suppose $f$ is strictly decreasing on $xge 1$, so any positive integer $n$ satisfies $f(n+1)leint_n^{n+1}f(x)dxle f(n)$. Summing, $sum_{nge 2}f(n)leint_1^infty f(x)dxlesum_{nge 1}f(n)$. Equivalently, $int_1^infty f(x)dxlesum_{nge 1}f(n)le f(1)+int_1^infty f(x)dx$. You just need the choice $f(x)=x^{-2}$. Another important corollary, called the integral test, is that for such $f$ the series $sum_{nge 1}f(n)$ converges iff $int_1^infty f(x)dx$ does. In particular, the divergence of the harmonic series is equivalent to $int_1^inftyfrac{dx}{x}=lninfty=infty$.






                                      share|cite|improve this answer












                                      More generally, suppose $f$ is strictly decreasing on $xge 1$, so any positive integer $n$ satisfies $f(n+1)leint_n^{n+1}f(x)dxle f(n)$. Summing, $sum_{nge 2}f(n)leint_1^infty f(x)dxlesum_{nge 1}f(n)$. Equivalently, $int_1^infty f(x)dxlesum_{nge 1}f(n)le f(1)+int_1^infty f(x)dx$. You just need the choice $f(x)=x^{-2}$. Another important corollary, called the integral test, is that for such $f$ the series $sum_{nge 1}f(n)$ converges iff $int_1^infty f(x)dx$ does. In particular, the divergence of the harmonic series is equivalent to $int_1^inftyfrac{dx}{x}=lninfty=infty$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Nov 27 '18 at 19:12









                                      J.G.

                                      22.6k22136




                                      22.6k22136






























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