Comparing integral and series of $1/(x^a)$












2












$begingroup$


The problem is to $$sum^N_{n=2}frac{1}{n^a}leqint^N_1frac{1}{x^a}$$ and to use this to prove the convergence of the series for $a>1$.
So, I believe I have the second part down. Namely evaluating the integral for the cases




  1. $$a>1Rightarrowint^N_1frac{1}{x^a}=frac{1}{1-a}$$

  2. $$a=1Rightarrowint^N_1frac{1}{x^a}=infty$$


  3. $$0<a<1Rightarrowint^N_1frac{1}{x^a}=infty$$
    and so using the [unproven fact] that $sum^N_{n=2}frac{1}{n^a}leqint^N_1frac{1}{x^a}$, the latter part follows from the Comparison Test. Still, I am having a hard time proving the former part of this question.






real-analysis integration sequences-and-series analysis definite-integrals







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$endgroup$

















    2












    $begingroup$


    The problem is to $$sum^N_{n=2}frac{1}{n^a}leqint^N_1frac{1}{x^a}$$ and to use this to prove the convergence of the series for $a>1$.
    So, I believe I have the second part down. Namely evaluating the integral for the cases




    1. $$a>1Rightarrowint^N_1frac{1}{x^a}=frac{1}{1-a}$$

    2. $$a=1Rightarrowint^N_1frac{1}{x^a}=infty$$


    3. $$0<a<1Rightarrowint^N_1frac{1}{x^a}=infty$$
      and so using the [unproven fact] that $sum^N_{n=2}frac{1}{n^a}leqint^N_1frac{1}{x^a}$, the latter part follows from the Comparison Test. Still, I am having a hard time proving the former part of this question.






    real-analysis integration sequences-and-series analysis definite-integrals







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      The problem is to $$sum^N_{n=2}frac{1}{n^a}leqint^N_1frac{1}{x^a}$$ and to use this to prove the convergence of the series for $a>1$.
      So, I believe I have the second part down. Namely evaluating the integral for the cases




      1. $$a>1Rightarrowint^N_1frac{1}{x^a}=frac{1}{1-a}$$

      2. $$a=1Rightarrowint^N_1frac{1}{x^a}=infty$$


      3. $$0<a<1Rightarrowint^N_1frac{1}{x^a}=infty$$
        and so using the [unproven fact] that $sum^N_{n=2}frac{1}{n^a}leqint^N_1frac{1}{x^a}$, the latter part follows from the Comparison Test. Still, I am having a hard time proving the former part of this question.






      real-analysis integration sequences-and-series analysis definite-integrals







      share|cite|improve this question











      $endgroup$




      The problem is to $$sum^N_{n=2}frac{1}{n^a}leqint^N_1frac{1}{x^a}$$ and to use this to prove the convergence of the series for $a>1$.
      So, I believe I have the second part down. Namely evaluating the integral for the cases




      1. $$a>1Rightarrowint^N_1frac{1}{x^a}=frac{1}{1-a}$$

      2. $$a=1Rightarrowint^N_1frac{1}{x^a}=infty$$


      3. $$0<a<1Rightarrowint^N_1frac{1}{x^a}=infty$$
        and so using the [unproven fact] that $sum^N_{n=2}frac{1}{n^a}leqint^N_1frac{1}{x^a}$, the latter part follows from the Comparison Test. Still, I am having a hard time proving the former part of this question.






      real-analysis integration sequences-and-series analysis definite-integrals




      real-analysis integration sequences-and-series analysis definite-integrals






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 10 '18 at 20:49









      Key Flex

      8,00261233




      8,00261233










      asked Dec 10 '18 at 20:36









      Gwen DiGwen Di

      648




      648






















          3 Answers
          3






          active

          oldest

          votes


















          0












          $begingroup$

          First notice that if $ninmathbb N$, since $1/x^agt 1/(n+1)^a$ for all $xin (n,n+1)$, we have that
          $$int_n^{n+1}frac{dx}{x^a}gt int_n^{n+1}frac{dx}{(n+1)^a}=frac{1}{(n+1)^a}$$
          This implies that
          $$begin{align}
          int_1^N frac{dx}{x^a} &=int_1^2 frac{dx}{x^a}+int_2^3 frac{dx}{x^a}+...+int_{N-1}^N frac{dx}{x^a}\
          &> frac{1}{2^a}+frac{1}{3^a}+...+frac{1}{N^a}\
          &=sum_{n=2}^N frac{1}{n^2}
          end{align}$$
          which proves the desired inequality:
          $$int_1^N frac{dx}{x^a}gt sum_{n=2}^N frac{1}{n^2}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thanks, but a couple questions: 1) not sure how you evaluated that second integral in the second line to be 1/(n+1)^a ... 2) I am also a bit confused on how the last inequality proven related to the inequality proposed in the question (namely, 1/n^a, not 1/n^2)
            $endgroup$
            – Gwen Di
            Dec 10 '18 at 22:41



















          0












          $begingroup$

          Make a sketch for $frac1{x^a}$ and $frac1{(x-1)^a}$



          enter image description here



          then we have



          $$sum^N_{n=2}frac{1}{n^a}leqint^{N+1}_2frac{1}{(x-1)^a}=int^N_1frac{1}{x^a}$$






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            We show that $${1over (k+1)^a}le int_{k}^{k+1} {1over x^a}dx$$and then by summing up the sided from $k=1$ to $k=N-1$ we are done. From the other side we know $${1over (k+1)^a}=int_k^{k+1} {dxover (lfloor xrfloor +1)^a}$$by the definition of floor function. Also $$lfloor xrfloorle x<lfloor xrfloor+1$$therefore for $xge 1$ and $a>1$ we obtain $$(lfloor xrfloor+1)^a>x^a$$or equivalently $${1over (lfloor xrfloor+1)^a}<{1over x^a}$$by integrating the both sides, we prove the integral inequality first, hence the general problem.






            share|cite|improve this answer









            $endgroup$













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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              0












              $begingroup$

              First notice that if $ninmathbb N$, since $1/x^agt 1/(n+1)^a$ for all $xin (n,n+1)$, we have that
              $$int_n^{n+1}frac{dx}{x^a}gt int_n^{n+1}frac{dx}{(n+1)^a}=frac{1}{(n+1)^a}$$
              This implies that
              $$begin{align}
              int_1^N frac{dx}{x^a} &=int_1^2 frac{dx}{x^a}+int_2^3 frac{dx}{x^a}+...+int_{N-1}^N frac{dx}{x^a}\
              &> frac{1}{2^a}+frac{1}{3^a}+...+frac{1}{N^a}\
              &=sum_{n=2}^N frac{1}{n^2}
              end{align}$$
              which proves the desired inequality:
              $$int_1^N frac{dx}{x^a}gt sum_{n=2}^N frac{1}{n^2}$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                thanks, but a couple questions: 1) not sure how you evaluated that second integral in the second line to be 1/(n+1)^a ... 2) I am also a bit confused on how the last inequality proven related to the inequality proposed in the question (namely, 1/n^a, not 1/n^2)
                $endgroup$
                – Gwen Di
                Dec 10 '18 at 22:41
















              0












              $begingroup$

              First notice that if $ninmathbb N$, since $1/x^agt 1/(n+1)^a$ for all $xin (n,n+1)$, we have that
              $$int_n^{n+1}frac{dx}{x^a}gt int_n^{n+1}frac{dx}{(n+1)^a}=frac{1}{(n+1)^a}$$
              This implies that
              $$begin{align}
              int_1^N frac{dx}{x^a} &=int_1^2 frac{dx}{x^a}+int_2^3 frac{dx}{x^a}+...+int_{N-1}^N frac{dx}{x^a}\
              &> frac{1}{2^a}+frac{1}{3^a}+...+frac{1}{N^a}\
              &=sum_{n=2}^N frac{1}{n^2}
              end{align}$$
              which proves the desired inequality:
              $$int_1^N frac{dx}{x^a}gt sum_{n=2}^N frac{1}{n^2}$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                thanks, but a couple questions: 1) not sure how you evaluated that second integral in the second line to be 1/(n+1)^a ... 2) I am also a bit confused on how the last inequality proven related to the inequality proposed in the question (namely, 1/n^a, not 1/n^2)
                $endgroup$
                – Gwen Di
                Dec 10 '18 at 22:41














              0












              0








              0





              $begingroup$

              First notice that if $ninmathbb N$, since $1/x^agt 1/(n+1)^a$ for all $xin (n,n+1)$, we have that
              $$int_n^{n+1}frac{dx}{x^a}gt int_n^{n+1}frac{dx}{(n+1)^a}=frac{1}{(n+1)^a}$$
              This implies that
              $$begin{align}
              int_1^N frac{dx}{x^a} &=int_1^2 frac{dx}{x^a}+int_2^3 frac{dx}{x^a}+...+int_{N-1}^N frac{dx}{x^a}\
              &> frac{1}{2^a}+frac{1}{3^a}+...+frac{1}{N^a}\
              &=sum_{n=2}^N frac{1}{n^2}
              end{align}$$
              which proves the desired inequality:
              $$int_1^N frac{dx}{x^a}gt sum_{n=2}^N frac{1}{n^2}$$






              share|cite|improve this answer









              $endgroup$



              First notice that if $ninmathbb N$, since $1/x^agt 1/(n+1)^a$ for all $xin (n,n+1)$, we have that
              $$int_n^{n+1}frac{dx}{x^a}gt int_n^{n+1}frac{dx}{(n+1)^a}=frac{1}{(n+1)^a}$$
              This implies that
              $$begin{align}
              int_1^N frac{dx}{x^a} &=int_1^2 frac{dx}{x^a}+int_2^3 frac{dx}{x^a}+...+int_{N-1}^N frac{dx}{x^a}\
              &> frac{1}{2^a}+frac{1}{3^a}+...+frac{1}{N^a}\
              &=sum_{n=2}^N frac{1}{n^2}
              end{align}$$
              which proves the desired inequality:
              $$int_1^N frac{dx}{x^a}gt sum_{n=2}^N frac{1}{n^2}$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 10 '18 at 20:47









              FrpzzdFrpzzd

              22.9k841109




              22.9k841109












              • $begingroup$
                thanks, but a couple questions: 1) not sure how you evaluated that second integral in the second line to be 1/(n+1)^a ... 2) I am also a bit confused on how the last inequality proven related to the inequality proposed in the question (namely, 1/n^a, not 1/n^2)
                $endgroup$
                – Gwen Di
                Dec 10 '18 at 22:41


















              • $begingroup$
                thanks, but a couple questions: 1) not sure how you evaluated that second integral in the second line to be 1/(n+1)^a ... 2) I am also a bit confused on how the last inequality proven related to the inequality proposed in the question (namely, 1/n^a, not 1/n^2)
                $endgroup$
                – Gwen Di
                Dec 10 '18 at 22:41
















              $begingroup$
              thanks, but a couple questions: 1) not sure how you evaluated that second integral in the second line to be 1/(n+1)^a ... 2) I am also a bit confused on how the last inequality proven related to the inequality proposed in the question (namely, 1/n^a, not 1/n^2)
              $endgroup$
              – Gwen Di
              Dec 10 '18 at 22:41




              $begingroup$
              thanks, but a couple questions: 1) not sure how you evaluated that second integral in the second line to be 1/(n+1)^a ... 2) I am also a bit confused on how the last inequality proven related to the inequality proposed in the question (namely, 1/n^a, not 1/n^2)
              $endgroup$
              – Gwen Di
              Dec 10 '18 at 22:41











              0












              $begingroup$

              Make a sketch for $frac1{x^a}$ and $frac1{(x-1)^a}$



              enter image description here



              then we have



              $$sum^N_{n=2}frac{1}{n^a}leqint^{N+1}_2frac{1}{(x-1)^a}=int^N_1frac{1}{x^a}$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Make a sketch for $frac1{x^a}$ and $frac1{(x-1)^a}$



                enter image description here



                then we have



                $$sum^N_{n=2}frac{1}{n^a}leqint^{N+1}_2frac{1}{(x-1)^a}=int^N_1frac{1}{x^a}$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Make a sketch for $frac1{x^a}$ and $frac1{(x-1)^a}$



                  enter image description here



                  then we have



                  $$sum^N_{n=2}frac{1}{n^a}leqint^{N+1}_2frac{1}{(x-1)^a}=int^N_1frac{1}{x^a}$$






                  share|cite|improve this answer









                  $endgroup$



                  Make a sketch for $frac1{x^a}$ and $frac1{(x-1)^a}$



                  enter image description here



                  then we have



                  $$sum^N_{n=2}frac{1}{n^a}leqint^{N+1}_2frac{1}{(x-1)^a}=int^N_1frac{1}{x^a}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 10 '18 at 20:51









                  gimusigimusi

                  92.8k84494




                  92.8k84494























                      0












                      $begingroup$

                      We show that $${1over (k+1)^a}le int_{k}^{k+1} {1over x^a}dx$$and then by summing up the sided from $k=1$ to $k=N-1$ we are done. From the other side we know $${1over (k+1)^a}=int_k^{k+1} {dxover (lfloor xrfloor +1)^a}$$by the definition of floor function. Also $$lfloor xrfloorle x<lfloor xrfloor+1$$therefore for $xge 1$ and $a>1$ we obtain $$(lfloor xrfloor+1)^a>x^a$$or equivalently $${1over (lfloor xrfloor+1)^a}<{1over x^a}$$by integrating the both sides, we prove the integral inequality first, hence the general problem.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        We show that $${1over (k+1)^a}le int_{k}^{k+1} {1over x^a}dx$$and then by summing up the sided from $k=1$ to $k=N-1$ we are done. From the other side we know $${1over (k+1)^a}=int_k^{k+1} {dxover (lfloor xrfloor +1)^a}$$by the definition of floor function. Also $$lfloor xrfloorle x<lfloor xrfloor+1$$therefore for $xge 1$ and $a>1$ we obtain $$(lfloor xrfloor+1)^a>x^a$$or equivalently $${1over (lfloor xrfloor+1)^a}<{1over x^a}$$by integrating the both sides, we prove the integral inequality first, hence the general problem.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          We show that $${1over (k+1)^a}le int_{k}^{k+1} {1over x^a}dx$$and then by summing up the sided from $k=1$ to $k=N-1$ we are done. From the other side we know $${1over (k+1)^a}=int_k^{k+1} {dxover (lfloor xrfloor +1)^a}$$by the definition of floor function. Also $$lfloor xrfloorle x<lfloor xrfloor+1$$therefore for $xge 1$ and $a>1$ we obtain $$(lfloor xrfloor+1)^a>x^a$$or equivalently $${1over (lfloor xrfloor+1)^a}<{1over x^a}$$by integrating the both sides, we prove the integral inequality first, hence the general problem.






                          share|cite|improve this answer









                          $endgroup$



                          We show that $${1over (k+1)^a}le int_{k}^{k+1} {1over x^a}dx$$and then by summing up the sided from $k=1$ to $k=N-1$ we are done. From the other side we know $${1over (k+1)^a}=int_k^{k+1} {dxover (lfloor xrfloor +1)^a}$$by the definition of floor function. Also $$lfloor xrfloorle x<lfloor xrfloor+1$$therefore for $xge 1$ and $a>1$ we obtain $$(lfloor xrfloor+1)^a>x^a$$or equivalently $${1over (lfloor xrfloor+1)^a}<{1over x^a}$$by integrating the both sides, we prove the integral inequality first, hence the general problem.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 15 '18 at 0:00









                          Mostafa AyazMostafa Ayaz

                          15.5k3939




                          15.5k3939






























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