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











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



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







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















up vote
2
down vote

favorite












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



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 at 18:47













up vote
2
down vote

favorite









up vote
2
down vote

favorite











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



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






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asked Nov 27 at 18:44









James Mitchell

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


















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
















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




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










4 Answers
4






active

oldest

votes

















up vote
2
down vote













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




























    up vote
    1
    down vote













    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




























      up vote
      1
      down vote













      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




























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        0
        down vote













        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





















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          4 Answers
          4






          active

          oldest

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          4 Answers
          4






          active

          oldest

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          active

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          active

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          up vote
          2
          down vote













          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

























            up vote
            2
            down vote













            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























              up vote
              2
              down vote










              up vote
              2
              down vote









              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 at 19:01









              rubikscube09

              1,167717




              1,167717






















                  up vote
                  1
                  down vote













                  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

























                    up vote
                    1
                    down vote













                    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























                      up vote
                      1
                      down vote










                      up vote
                      1
                      down vote









                      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 at 18:47









                      gimusi

                      92.3k84495




                      92.3k84495






















                          up vote
                          1
                          down vote













                          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

























                            up vote
                            1
                            down vote













                            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























                              up vote
                              1
                              down vote










                              up vote
                              1
                              down vote









                              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 at 19:09









                              zhw.

                              71.1k43075




                              71.1k43075






















                                  up vote
                                  0
                                  down vote













                                  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

























                                    up vote
                                    0
                                    down vote













                                    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























                                      up vote
                                      0
                                      down vote










                                      up vote
                                      0
                                      down vote









                                      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 at 19:12









                                      J.G.

                                      21.1k21933




                                      21.1k21933






























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