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












2














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





























    2














    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



























      2












      2








      2







      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















      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

      7,53441232




      7,53441232










      asked Dec 10 '18 at 20:36









      Gwen Di

      648




      648






















          3 Answers
          3






          active

          oldest

          votes


















          0














          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





















          • 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)
            – Gwen Di
            Dec 10 '18 at 22:41



















          0














          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





























            0














            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





















              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%2f3034429%2fcomparing-integral-and-series-of-1-xa%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              0














              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





















              • 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)
                – Gwen Di
                Dec 10 '18 at 22:41
















              0














              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





















              • 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)
                – Gwen Di
                Dec 10 '18 at 22:41














              0












              0








              0






              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












              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









              Frpzzd

              22.1k839107




              22.1k839107












              • 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)
                – Gwen Di
                Dec 10 '18 at 22:41


















              • 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)
                – Gwen Di
                Dec 10 '18 at 22:41
















              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)
              – Gwen Di
              Dec 10 '18 at 22:41




              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)
              – Gwen Di
              Dec 10 '18 at 22:41











              0














              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


























                0














                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
























                  0












                  0








                  0






                  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












                  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









                  gimusi

                  1




                  1























                      0














                      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


























                        0














                        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
























                          0












                          0








                          0






                          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












                          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 Ayaz

                          13.7k3936




                          13.7k3936






























                              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.





                              Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                              Please pay close attention to the following guidance:


                              • 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.


                              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%2f3034429%2fcomparing-integral-and-series-of-1-xa%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

                              Bressuire

                              Image
                              .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                              Read more

                              Cabo Verde

                              Image
                                Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
                              Read more

                              Gyllenstierna

                              Image
                              .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                              Read more