$lim_{ntoinfty}left(sum_{k=0}^{n-1}(zeta(2)-H_{k,2})-H_nright)=1$











up vote
5
down vote

favorite
3












I found this limit in a book, without any explanation:



$$lim_{ntoinfty}left(sum_{k=0}^{n-1}(zeta(2)-H_{k,2})-H_nright)=1$$



where $H_{k,2}:=sum_{j=1}^kfrac1{j^2}$. However Im unable to find the value of this limit from myself. After some work I get the equivalent expression



$$lim_{ntoinfty}sum_{k=0}^{n-1}sum_{j=k}^inftyfrac1{(j+1)^2(j+2)}$$



but anyway Im stuck here. Can someone show me a way to compute this limit? Thank you.



UPDATE: Wolfram Mathematica computed it value perfectly, so I guess there is some integral or algebraic identity from where to calculate it.










share|cite|improve this question




























    up vote
    5
    down vote

    favorite
    3












    I found this limit in a book, without any explanation:



    $$lim_{ntoinfty}left(sum_{k=0}^{n-1}(zeta(2)-H_{k,2})-H_nright)=1$$



    where $H_{k,2}:=sum_{j=1}^kfrac1{j^2}$. However Im unable to find the value of this limit from myself. After some work I get the equivalent expression



    $$lim_{ntoinfty}sum_{k=0}^{n-1}sum_{j=k}^inftyfrac1{(j+1)^2(j+2)}$$



    but anyway Im stuck here. Can someone show me a way to compute this limit? Thank you.



    UPDATE: Wolfram Mathematica computed it value perfectly, so I guess there is some integral or algebraic identity from where to calculate it.










    share|cite|improve this question


























      up vote
      5
      down vote

      favorite
      3









      up vote
      5
      down vote

      favorite
      3






      3





      I found this limit in a book, without any explanation:



      $$lim_{ntoinfty}left(sum_{k=0}^{n-1}(zeta(2)-H_{k,2})-H_nright)=1$$



      where $H_{k,2}:=sum_{j=1}^kfrac1{j^2}$. However Im unable to find the value of this limit from myself. After some work I get the equivalent expression



      $$lim_{ntoinfty}sum_{k=0}^{n-1}sum_{j=k}^inftyfrac1{(j+1)^2(j+2)}$$



      but anyway Im stuck here. Can someone show me a way to compute this limit? Thank you.



      UPDATE: Wolfram Mathematica computed it value perfectly, so I guess there is some integral or algebraic identity from where to calculate it.










      share|cite|improve this question















      I found this limit in a book, without any explanation:



      $$lim_{ntoinfty}left(sum_{k=0}^{n-1}(zeta(2)-H_{k,2})-H_nright)=1$$



      where $H_{k,2}:=sum_{j=1}^kfrac1{j^2}$. However Im unable to find the value of this limit from myself. After some work I get the equivalent expression



      $$lim_{ntoinfty}sum_{k=0}^{n-1}sum_{j=k}^inftyfrac1{(j+1)^2(j+2)}$$



      but anyway Im stuck here. Can someone show me a way to compute this limit? Thank you.



      UPDATE: Wolfram Mathematica computed it value perfectly, so I guess there is some integral or algebraic identity from where to calculate it.







      calculus limits special-functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 17 at 23:20

























      asked Nov 17 at 22:05









      Masacroso

      12.2k41746




      12.2k41746






















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          3
          down vote



          accepted










          Let's see:



          $$begin{eqnarray*} sum_{k=0}^{n-1}left(zeta(2)-H_k^{(2)}right) &=& zeta(2)+sum_{k=1}^{n-1}left(zeta(2)-H_k^{(2)}right)\&stackrel{text{SBP}}{=}&zeta(2)+(n-1)(zeta(2)-H_{n-1}^{(2)})+sum_{k=1}^{n-2}frac{k}{(k+1)^2}\&=&zeta(2)+(n-1)(zeta(2)-H_{n-1}^{(2)})+(H_{n-1}-1)-sum_{k=1}^{n-2}frac{1}{(k+1)^2}\&=&n( zeta(2)-H_{n-1}^{(2)})+H_{n-1}end{eqnarray*}$$
          hence the claim is equivalent to



          $$ lim_{nto +infty} n(zeta(2)-H_{n-1}^{(2)}) = lim_{nto +infty}nsum_{mgeq n}frac{1}{m^2} = 1 $$
          which is pretty clear since $sum_{mgeq n}frac{1}{m^2} = Oleft(frac{1}{n^2}right)+int_{n}^{+infty}frac{dx}{x^2}=frac{1}{n}+Oleft(frac{1}{n^2}right)$.
          $text{SBP}$ stands for Summation By Parts, of course.






          share|cite|improve this answer




























            up vote
            2
            down vote













            Considering your last expression $$a_n=sum_{k=0}^{n-1}sum_{j=k}^inftyfrac1{(j+1)^2(j+2)}$$
            $$sum_{j=k}^inftyfrac1{(j+1)^2(j+2)}=psi ^{(1)}(k+1)-frac{1}{k+1}$$ making
            $$a_n=n ,psi ^{(1)}(n+1)$$ the expansion of which being
            $$a_n=1-frac{1}{2 n}+frac{1}{6 n^2}+Oleft(frac{1}{n^4}right)$$






            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',
              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%2f3002874%2flim-n-to-infty-left-sum-k-0n-1-zeta2-h-k-2-h-n-right-1%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              3
              down vote



              accepted










              Let's see:



              $$begin{eqnarray*} sum_{k=0}^{n-1}left(zeta(2)-H_k^{(2)}right) &=& zeta(2)+sum_{k=1}^{n-1}left(zeta(2)-H_k^{(2)}right)\&stackrel{text{SBP}}{=}&zeta(2)+(n-1)(zeta(2)-H_{n-1}^{(2)})+sum_{k=1}^{n-2}frac{k}{(k+1)^2}\&=&zeta(2)+(n-1)(zeta(2)-H_{n-1}^{(2)})+(H_{n-1}-1)-sum_{k=1}^{n-2}frac{1}{(k+1)^2}\&=&n( zeta(2)-H_{n-1}^{(2)})+H_{n-1}end{eqnarray*}$$
              hence the claim is equivalent to



              $$ lim_{nto +infty} n(zeta(2)-H_{n-1}^{(2)}) = lim_{nto +infty}nsum_{mgeq n}frac{1}{m^2} = 1 $$
              which is pretty clear since $sum_{mgeq n}frac{1}{m^2} = Oleft(frac{1}{n^2}right)+int_{n}^{+infty}frac{dx}{x^2}=frac{1}{n}+Oleft(frac{1}{n^2}right)$.
              $text{SBP}$ stands for Summation By Parts, of course.






              share|cite|improve this answer

























                up vote
                3
                down vote



                accepted










                Let's see:



                $$begin{eqnarray*} sum_{k=0}^{n-1}left(zeta(2)-H_k^{(2)}right) &=& zeta(2)+sum_{k=1}^{n-1}left(zeta(2)-H_k^{(2)}right)\&stackrel{text{SBP}}{=}&zeta(2)+(n-1)(zeta(2)-H_{n-1}^{(2)})+sum_{k=1}^{n-2}frac{k}{(k+1)^2}\&=&zeta(2)+(n-1)(zeta(2)-H_{n-1}^{(2)})+(H_{n-1}-1)-sum_{k=1}^{n-2}frac{1}{(k+1)^2}\&=&n( zeta(2)-H_{n-1}^{(2)})+H_{n-1}end{eqnarray*}$$
                hence the claim is equivalent to



                $$ lim_{nto +infty} n(zeta(2)-H_{n-1}^{(2)}) = lim_{nto +infty}nsum_{mgeq n}frac{1}{m^2} = 1 $$
                which is pretty clear since $sum_{mgeq n}frac{1}{m^2} = Oleft(frac{1}{n^2}right)+int_{n}^{+infty}frac{dx}{x^2}=frac{1}{n}+Oleft(frac{1}{n^2}right)$.
                $text{SBP}$ stands for Summation By Parts, of course.






                share|cite|improve this answer























                  up vote
                  3
                  down vote



                  accepted







                  up vote
                  3
                  down vote



                  accepted






                  Let's see:



                  $$begin{eqnarray*} sum_{k=0}^{n-1}left(zeta(2)-H_k^{(2)}right) &=& zeta(2)+sum_{k=1}^{n-1}left(zeta(2)-H_k^{(2)}right)\&stackrel{text{SBP}}{=}&zeta(2)+(n-1)(zeta(2)-H_{n-1}^{(2)})+sum_{k=1}^{n-2}frac{k}{(k+1)^2}\&=&zeta(2)+(n-1)(zeta(2)-H_{n-1}^{(2)})+(H_{n-1}-1)-sum_{k=1}^{n-2}frac{1}{(k+1)^2}\&=&n( zeta(2)-H_{n-1}^{(2)})+H_{n-1}end{eqnarray*}$$
                  hence the claim is equivalent to



                  $$ lim_{nto +infty} n(zeta(2)-H_{n-1}^{(2)}) = lim_{nto +infty}nsum_{mgeq n}frac{1}{m^2} = 1 $$
                  which is pretty clear since $sum_{mgeq n}frac{1}{m^2} = Oleft(frac{1}{n^2}right)+int_{n}^{+infty}frac{dx}{x^2}=frac{1}{n}+Oleft(frac{1}{n^2}right)$.
                  $text{SBP}$ stands for Summation By Parts, of course.






                  share|cite|improve this answer












                  Let's see:



                  $$begin{eqnarray*} sum_{k=0}^{n-1}left(zeta(2)-H_k^{(2)}right) &=& zeta(2)+sum_{k=1}^{n-1}left(zeta(2)-H_k^{(2)}right)\&stackrel{text{SBP}}{=}&zeta(2)+(n-1)(zeta(2)-H_{n-1}^{(2)})+sum_{k=1}^{n-2}frac{k}{(k+1)^2}\&=&zeta(2)+(n-1)(zeta(2)-H_{n-1}^{(2)})+(H_{n-1}-1)-sum_{k=1}^{n-2}frac{1}{(k+1)^2}\&=&n( zeta(2)-H_{n-1}^{(2)})+H_{n-1}end{eqnarray*}$$
                  hence the claim is equivalent to



                  $$ lim_{nto +infty} n(zeta(2)-H_{n-1}^{(2)}) = lim_{nto +infty}nsum_{mgeq n}frac{1}{m^2} = 1 $$
                  which is pretty clear since $sum_{mgeq n}frac{1}{m^2} = Oleft(frac{1}{n^2}right)+int_{n}^{+infty}frac{dx}{x^2}=frac{1}{n}+Oleft(frac{1}{n^2}right)$.
                  $text{SBP}$ stands for Summation By Parts, of course.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 17 at 23:23









                  Jack D'Aurizio

                  283k33275653




                  283k33275653






















                      up vote
                      2
                      down vote













                      Considering your last expression $$a_n=sum_{k=0}^{n-1}sum_{j=k}^inftyfrac1{(j+1)^2(j+2)}$$
                      $$sum_{j=k}^inftyfrac1{(j+1)^2(j+2)}=psi ^{(1)}(k+1)-frac{1}{k+1}$$ making
                      $$a_n=n ,psi ^{(1)}(n+1)$$ the expansion of which being
                      $$a_n=1-frac{1}{2 n}+frac{1}{6 n^2}+Oleft(frac{1}{n^4}right)$$






                      share|cite|improve this answer

























                        up vote
                        2
                        down vote













                        Considering your last expression $$a_n=sum_{k=0}^{n-1}sum_{j=k}^inftyfrac1{(j+1)^2(j+2)}$$
                        $$sum_{j=k}^inftyfrac1{(j+1)^2(j+2)}=psi ^{(1)}(k+1)-frac{1}{k+1}$$ making
                        $$a_n=n ,psi ^{(1)}(n+1)$$ the expansion of which being
                        $$a_n=1-frac{1}{2 n}+frac{1}{6 n^2}+Oleft(frac{1}{n^4}right)$$






                        share|cite|improve this answer























                          up vote
                          2
                          down vote










                          up vote
                          2
                          down vote









                          Considering your last expression $$a_n=sum_{k=0}^{n-1}sum_{j=k}^inftyfrac1{(j+1)^2(j+2)}$$
                          $$sum_{j=k}^inftyfrac1{(j+1)^2(j+2)}=psi ^{(1)}(k+1)-frac{1}{k+1}$$ making
                          $$a_n=n ,psi ^{(1)}(n+1)$$ the expansion of which being
                          $$a_n=1-frac{1}{2 n}+frac{1}{6 n^2}+Oleft(frac{1}{n^4}right)$$






                          share|cite|improve this answer












                          Considering your last expression $$a_n=sum_{k=0}^{n-1}sum_{j=k}^inftyfrac1{(j+1)^2(j+2)}$$
                          $$sum_{j=k}^inftyfrac1{(j+1)^2(j+2)}=psi ^{(1)}(k+1)-frac{1}{k+1}$$ making
                          $$a_n=n ,psi ^{(1)}(n+1)$$ the expansion of which being
                          $$a_n=1-frac{1}{2 n}+frac{1}{6 n^2}+Oleft(frac{1}{n^4}right)$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 18 at 6:06









                          Claude Leibovici

                          116k1156131




                          116k1156131






























                               

                              draft saved


                              draft discarded



















































                               


                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3002874%2flim-n-to-infty-left-sum-k-0n-1-zeta2-h-k-2-h-n-right-1%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

                              Ellipse (mathématiques)

                              Quarter-circle Tiles

                              Mont Emei