Limit of $x_n=sum_{k=np+1}^{nq}frac{1}{k}$ using Riemann sum












1












$begingroup$


I am trying to find the limit of the following sequence using Riemann sum:
$$x_n=sum_{k=np+1}^{nq}frac{1}{k}qquad p,qinmathbb{N}quad p<q$$
I have tried to develope the expression:
$$frac{1}{np+1}+frac{1}{np+2}+...+frac{1}{nq}=frac{1}{n}(frac{1}{p+frac{1}{n}}+frac{1}{p+frac{2}{n}}+...+frac{1}{p+frac{n(q-p)}{n}})=frac{1}{n}sum_{k=1}^{n(q-p)}frac{1}{p+frac{k}{n}}$$
But I need an expression like $frac{1}{n}sum_{k=1}^{n}f(frac{k}{n})$, from $k=1$ to $n$, not to $n(q-p)$, so I can calculate the limit as $int_0^1f(x)dx.$



Could you give me some hints? Thanks!






sequences-and-series limits riemann-sum







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I am trying to find the limit of the following sequence using Riemann sum:
    $$x_n=sum_{k=np+1}^{nq}frac{1}{k}qquad p,qinmathbb{N}quad p<q$$
    I have tried to develope the expression:
    $$frac{1}{np+1}+frac{1}{np+2}+...+frac{1}{nq}=frac{1}{n}(frac{1}{p+frac{1}{n}}+frac{1}{p+frac{2}{n}}+...+frac{1}{p+frac{n(q-p)}{n}})=frac{1}{n}sum_{k=1}^{n(q-p)}frac{1}{p+frac{k}{n}}$$
    But I need an expression like $frac{1}{n}sum_{k=1}^{n}f(frac{k}{n})$, from $k=1$ to $n$, not to $n(q-p)$, so I can calculate the limit as $int_0^1f(x)dx.$



    Could you give me some hints? Thanks!






    sequences-and-series limits riemann-sum







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      2



      $begingroup$


      I am trying to find the limit of the following sequence using Riemann sum:
      $$x_n=sum_{k=np+1}^{nq}frac{1}{k}qquad p,qinmathbb{N}quad p<q$$
      I have tried to develope the expression:
      $$frac{1}{np+1}+frac{1}{np+2}+...+frac{1}{nq}=frac{1}{n}(frac{1}{p+frac{1}{n}}+frac{1}{p+frac{2}{n}}+...+frac{1}{p+frac{n(q-p)}{n}})=frac{1}{n}sum_{k=1}^{n(q-p)}frac{1}{p+frac{k}{n}}$$
      But I need an expression like $frac{1}{n}sum_{k=1}^{n}f(frac{k}{n})$, from $k=1$ to $n$, not to $n(q-p)$, so I can calculate the limit as $int_0^1f(x)dx.$



      Could you give me some hints? Thanks!






      sequences-and-series limits riemann-sum







      share|cite|improve this question









      $endgroup$




      I am trying to find the limit of the following sequence using Riemann sum:
      $$x_n=sum_{k=np+1}^{nq}frac{1}{k}qquad p,qinmathbb{N}quad p<q$$
      I have tried to develope the expression:
      $$frac{1}{np+1}+frac{1}{np+2}+...+frac{1}{nq}=frac{1}{n}(frac{1}{p+frac{1}{n}}+frac{1}{p+frac{2}{n}}+...+frac{1}{p+frac{n(q-p)}{n}})=frac{1}{n}sum_{k=1}^{n(q-p)}frac{1}{p+frac{k}{n}}$$
      But I need an expression like $frac{1}{n}sum_{k=1}^{n}f(frac{k}{n})$, from $k=1$ to $n$, not to $n(q-p)$, so I can calculate the limit as $int_0^1f(x)dx.$



      Could you give me some hints? Thanks!






      sequences-and-series limits riemann-sum




      sequences-and-series limits riemann-sum






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 4 at 15:13









      GibbsGibbs

      137111




      137111






















          3 Answers
          3






          active

          oldest

          votes


















          0












          $begingroup$

          We just have to calculate $int_0^{q-p}$ instead of $int_0^1$. Then, with
          $$f(x)=frac{1}{p+x}$$
          $$int_0^{q-p}frac{1}{p+x}dx=[ln|p+x|]_0^{q-p}=ln|p+q-p|-ln|p|=lnfrac{q}{p}$$
          And we are done.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            $$lim_{nto infty}x_{n}{=lim_{nto infty}sum_{k=np}^{nq+1}{1over k}\=lim_{nto infty}sum_{k=np}^{nq+1}{nover k}cdot {1over n}\=lim_{nto infty}sum_{k=np}^{nq+1}{1over{kover n}}cdot {1over n}\=int_p^q{dxover x}\=ln {qover p}}$$






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              Some additional transformation have indeed to be made: you got
              $x_n=frac{1}{n}sum_{k=1}^{n(q-p)}frac{1}{p+frac{k}{n}}$; we can continue by writing
              $$
              x_n=frac{q-p}{nleft(q-pright)}sum_{k=1}^{n(q-p)}frac{1}{p+frac{kleft(q-pright)}{nleft(q-pright)}}= left(q-pright)frac 1{N_n}sum_{k=1}^{N_n}fleft(frac{k}{N_n}right)
              $$
              where $N_n=n (q-p)$ and $fcolon xmapsto 1/left(p+x(q-p)right)$.






              share|cite|improve this answer









              $endgroup$














                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%2f3061746%2flimit-of-x-n-sum-k-np1nq-frac1k-using-riemann-sum%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












                $begingroup$

                We just have to calculate $int_0^{q-p}$ instead of $int_0^1$. Then, with
                $$f(x)=frac{1}{p+x}$$
                $$int_0^{q-p}frac{1}{p+x}dx=[ln|p+x|]_0^{q-p}=ln|p+q-p|-ln|p|=lnfrac{q}{p}$$
                And we are done.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  We just have to calculate $int_0^{q-p}$ instead of $int_0^1$. Then, with
                  $$f(x)=frac{1}{p+x}$$
                  $$int_0^{q-p}frac{1}{p+x}dx=[ln|p+x|]_0^{q-p}=ln|p+q-p|-ln|p|=lnfrac{q}{p}$$
                  And we are done.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    We just have to calculate $int_0^{q-p}$ instead of $int_0^1$. Then, with
                    $$f(x)=frac{1}{p+x}$$
                    $$int_0^{q-p}frac{1}{p+x}dx=[ln|p+x|]_0^{q-p}=ln|p+q-p|-ln|p|=lnfrac{q}{p}$$
                    And we are done.






                    share|cite|improve this answer









                    $endgroup$



                    We just have to calculate $int_0^{q-p}$ instead of $int_0^1$. Then, with
                    $$f(x)=frac{1}{p+x}$$
                    $$int_0^{q-p}frac{1}{p+x}dx=[ln|p+x|]_0^{q-p}=ln|p+q-p|-ln|p|=lnfrac{q}{p}$$
                    And we are done.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 4 at 15:45









                    GibbsGibbs

                    137111




                    137111























                        0












                        $begingroup$

                        $$lim_{nto infty}x_{n}{=lim_{nto infty}sum_{k=np}^{nq+1}{1over k}\=lim_{nto infty}sum_{k=np}^{nq+1}{nover k}cdot {1over n}\=lim_{nto infty}sum_{k=np}^{nq+1}{1over{kover n}}cdot {1over n}\=int_p^q{dxover x}\=ln {qover p}}$$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          $$lim_{nto infty}x_{n}{=lim_{nto infty}sum_{k=np}^{nq+1}{1over k}\=lim_{nto infty}sum_{k=np}^{nq+1}{nover k}cdot {1over n}\=lim_{nto infty}sum_{k=np}^{nq+1}{1over{kover n}}cdot {1over n}\=int_p^q{dxover x}\=ln {qover p}}$$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            $$lim_{nto infty}x_{n}{=lim_{nto infty}sum_{k=np}^{nq+1}{1over k}\=lim_{nto infty}sum_{k=np}^{nq+1}{nover k}cdot {1over n}\=lim_{nto infty}sum_{k=np}^{nq+1}{1over{kover n}}cdot {1over n}\=int_p^q{dxover x}\=ln {qover p}}$$






                            share|cite|improve this answer









                            $endgroup$



                            $$lim_{nto infty}x_{n}{=lim_{nto infty}sum_{k=np}^{nq+1}{1over k}\=lim_{nto infty}sum_{k=np}^{nq+1}{nover k}cdot {1over n}\=lim_{nto infty}sum_{k=np}^{nq+1}{1over{kover n}}cdot {1over n}\=int_p^q{dxover x}\=ln {qover p}}$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 5 at 6:57









                            Mostafa AyazMostafa Ayaz

                            18.2k31040




                            18.2k31040























                                0












                                $begingroup$

                                Some additional transformation have indeed to be made: you got
                                $x_n=frac{1}{n}sum_{k=1}^{n(q-p)}frac{1}{p+frac{k}{n}}$; we can continue by writing
                                $$
                                x_n=frac{q-p}{nleft(q-pright)}sum_{k=1}^{n(q-p)}frac{1}{p+frac{kleft(q-pright)}{nleft(q-pright)}}= left(q-pright)frac 1{N_n}sum_{k=1}^{N_n}fleft(frac{k}{N_n}right)
                                $$
                                where $N_n=n (q-p)$ and $fcolon xmapsto 1/left(p+x(q-p)right)$.






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  Some additional transformation have indeed to be made: you got
                                  $x_n=frac{1}{n}sum_{k=1}^{n(q-p)}frac{1}{p+frac{k}{n}}$; we can continue by writing
                                  $$
                                  x_n=frac{q-p}{nleft(q-pright)}sum_{k=1}^{n(q-p)}frac{1}{p+frac{kleft(q-pright)}{nleft(q-pright)}}= left(q-pright)frac 1{N_n}sum_{k=1}^{N_n}fleft(frac{k}{N_n}right)
                                  $$
                                  where $N_n=n (q-p)$ and $fcolon xmapsto 1/left(p+x(q-p)right)$.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    Some additional transformation have indeed to be made: you got
                                    $x_n=frac{1}{n}sum_{k=1}^{n(q-p)}frac{1}{p+frac{k}{n}}$; we can continue by writing
                                    $$
                                    x_n=frac{q-p}{nleft(q-pright)}sum_{k=1}^{n(q-p)}frac{1}{p+frac{kleft(q-pright)}{nleft(q-pright)}}= left(q-pright)frac 1{N_n}sum_{k=1}^{N_n}fleft(frac{k}{N_n}right)
                                    $$
                                    where $N_n=n (q-p)$ and $fcolon xmapsto 1/left(p+x(q-p)right)$.






                                    share|cite|improve this answer









                                    $endgroup$



                                    Some additional transformation have indeed to be made: you got
                                    $x_n=frac{1}{n}sum_{k=1}^{n(q-p)}frac{1}{p+frac{k}{n}}$; we can continue by writing
                                    $$
                                    x_n=frac{q-p}{nleft(q-pright)}sum_{k=1}^{n(q-p)}frac{1}{p+frac{kleft(q-pright)}{nleft(q-pright)}}= left(q-pright)frac 1{N_n}sum_{k=1}^{N_n}fleft(frac{k}{N_n}right)
                                    $$
                                    where $N_n=n (q-p)$ and $fcolon xmapsto 1/left(p+x(q-p)right)$.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Jan 5 at 13:16









                                    Davide GiraudoDavide Giraudo

                                    128k17156268




                                    128k17156268






























                                        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.




                                        draft saved


                                        draft discarded














                                        StackExchange.ready(
                                        function () {
                                        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061746%2flimit-of-x-n-sum-k-np1nq-frac1k-using-riemann-sum%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

                                        Probability when a professor distributes a quiz and homework assignment to a class of n students.

                                        Image
                                        0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
                                        Read more

                                        Aardman Animations

                                        Image
                                        Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
                                        Read more

                                        Are they similar matrix

                                        Image
                                        0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
                                        Read more