Rearrangement of summations in series expansion of $-log(zeta(s))$












1












$begingroup$


This is a step in Titchmarsh's The theory of Riemann Zeta functions pg 2 of Chpt 1 eq 1.13 which I do not find obvious, though I can prove it directly. Assume $sigma=Re(s)>1$.



Denote by $pi(x)$ the number of primes $leq x$. Consider the following summation which is basically $-log(zeta(s))$.



begin{align} sum_{ngeq 2}a_n &:=sum_{ngeq 2}(pi(n)-pi(n-1))logleft(1-tfrac{1}{n^s}right) \ &=sum_{ngeq 2}pi(n)left(logleft(1-tfrac{1}{n^s}right)-logleft(1-tfrac{1}{(n+1)^s}right)right)\&:=sum_{ngeq 2}b_nend{align} Rearrangement is allowed because $log(1-n^{-s})sim n^{-sigma}$ and $pi(n)leq n $



$textbf{Q:}$ Why is this obvious? I considered $$|a_n-b_n|=left|-pi(n-1)logleft(1-tfrac{1}{n^s}right)+pi(n)logleft(1-tfrac{1}{(n+1)^s}right)right|.$$ Add and subtract term $pi(n-1)log(1-frac{1}{(n+1)^s})$ and use triangle inequality to see $$|a_n-b_n|leq n^{-sigma}+left|pi(n-1)left(logleft(1-tfrac{1}{(1+n)^s}right)-logleft(1-tfrac{1}{n^s}right)right)right|.$$ The second term is bounded by order $n^{-2sigma}$. Hence I have tail controlled. However this does not directly yield equality but says tail can be arbitrarily controlled.






real-analysis complex-analysis number-theory riemann-zeta







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    This is a step in Titchmarsh's The theory of Riemann Zeta functions pg 2 of Chpt 1 eq 1.13 which I do not find obvious, though I can prove it directly. Assume $sigma=Re(s)>1$.



    Denote by $pi(x)$ the number of primes $leq x$. Consider the following summation which is basically $-log(zeta(s))$.



    begin{align} sum_{ngeq 2}a_n &:=sum_{ngeq 2}(pi(n)-pi(n-1))logleft(1-tfrac{1}{n^s}right) \ &=sum_{ngeq 2}pi(n)left(logleft(1-tfrac{1}{n^s}right)-logleft(1-tfrac{1}{(n+1)^s}right)right)\&:=sum_{ngeq 2}b_nend{align} Rearrangement is allowed because $log(1-n^{-s})sim n^{-sigma}$ and $pi(n)leq n $



    $textbf{Q:}$ Why is this obvious? I considered $$|a_n-b_n|=left|-pi(n-1)logleft(1-tfrac{1}{n^s}right)+pi(n)logleft(1-tfrac{1}{(n+1)^s}right)right|.$$ Add and subtract term $pi(n-1)log(1-frac{1}{(n+1)^s})$ and use triangle inequality to see $$|a_n-b_n|leq n^{-sigma}+left|pi(n-1)left(logleft(1-tfrac{1}{(1+n)^s}right)-logleft(1-tfrac{1}{n^s}right)right)right|.$$ The second term is bounded by order $n^{-2sigma}$. Hence I have tail controlled. However this does not directly yield equality but says tail can be arbitrarily controlled.






    real-analysis complex-analysis number-theory riemann-zeta







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      This is a step in Titchmarsh's The theory of Riemann Zeta functions pg 2 of Chpt 1 eq 1.13 which I do not find obvious, though I can prove it directly. Assume $sigma=Re(s)>1$.



      Denote by $pi(x)$ the number of primes $leq x$. Consider the following summation which is basically $-log(zeta(s))$.



      begin{align} sum_{ngeq 2}a_n &:=sum_{ngeq 2}(pi(n)-pi(n-1))logleft(1-tfrac{1}{n^s}right) \ &=sum_{ngeq 2}pi(n)left(logleft(1-tfrac{1}{n^s}right)-logleft(1-tfrac{1}{(n+1)^s}right)right)\&:=sum_{ngeq 2}b_nend{align} Rearrangement is allowed because $log(1-n^{-s})sim n^{-sigma}$ and $pi(n)leq n $



      $textbf{Q:}$ Why is this obvious? I considered $$|a_n-b_n|=left|-pi(n-1)logleft(1-tfrac{1}{n^s}right)+pi(n)logleft(1-tfrac{1}{(n+1)^s}right)right|.$$ Add and subtract term $pi(n-1)log(1-frac{1}{(n+1)^s})$ and use triangle inequality to see $$|a_n-b_n|leq n^{-sigma}+left|pi(n-1)left(logleft(1-tfrac{1}{(1+n)^s}right)-logleft(1-tfrac{1}{n^s}right)right)right|.$$ The second term is bounded by order $n^{-2sigma}$. Hence I have tail controlled. However this does not directly yield equality but says tail can be arbitrarily controlled.






      real-analysis complex-analysis number-theory riemann-zeta







      share|cite|improve this question











      $endgroup$




      This is a step in Titchmarsh's The theory of Riemann Zeta functions pg 2 of Chpt 1 eq 1.13 which I do not find obvious, though I can prove it directly. Assume $sigma=Re(s)>1$.



      Denote by $pi(x)$ the number of primes $leq x$. Consider the following summation which is basically $-log(zeta(s))$.



      begin{align} sum_{ngeq 2}a_n &:=sum_{ngeq 2}(pi(n)-pi(n-1))logleft(1-tfrac{1}{n^s}right) \ &=sum_{ngeq 2}pi(n)left(logleft(1-tfrac{1}{n^s}right)-logleft(1-tfrac{1}{(n+1)^s}right)right)\&:=sum_{ngeq 2}b_nend{align} Rearrangement is allowed because $log(1-n^{-s})sim n^{-sigma}$ and $pi(n)leq n $



      $textbf{Q:}$ Why is this obvious? I considered $$|a_n-b_n|=left|-pi(n-1)logleft(1-tfrac{1}{n^s}right)+pi(n)logleft(1-tfrac{1}{(n+1)^s}right)right|.$$ Add and subtract term $pi(n-1)log(1-frac{1}{(n+1)^s})$ and use triangle inequality to see $$|a_n-b_n|leq n^{-sigma}+left|pi(n-1)left(logleft(1-tfrac{1}{(1+n)^s}right)-logleft(1-tfrac{1}{n^s}right)right)right|.$$ The second term is bounded by order $n^{-2sigma}$. Hence I have tail controlled. However this does not directly yield equality but says tail can be arbitrarily controlled.






      real-analysis complex-analysis number-theory riemann-zeta




      real-analysis complex-analysis number-theory riemann-zeta






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 27 '18 at 1:56









      Kenny Wong

      19.1k21441




      19.1k21441










      asked Dec 27 '18 at 0:52









      user45765user45765

      2,6702724




      2,6702724






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          The point is that
          $$ left| sum_{n = 2}^N b_n - sum_{n = 2}^N a_n right| = left| pi (N) log left(1 - tfrac{1}{(N + 1)^s} right) right| leq CN^{1 - sigma}$$
          for some constant $C > 0$, and for sufficient large $N$.



          Since $1 - sigma < 0$, we have
          $$ lim_{N to infty} left| sum_{n = 2}^N b_n - sum_{n = 2}^N a_n right| = 0.$$



          So since $sum_{n = 2}^infty a_n$ converges, it must be true that $sum_{n = 2}^infty b_n$ converges too, and
          $ sum_{n = 2}^infty b_n = sum_{n = 2}^infty a_n$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Maybe this is a dumb question. Was this arrangement obvious at first sight? It is clear that it is natural to manipulate sum in that manner though it might not converge.
            $endgroup$
            – user45765
            Dec 27 '18 at 1:20










          • $begingroup$
            @user45765 I had to stare at it for a bit...
            $endgroup$
            – Kenny Wong
            Dec 27 '18 at 1:21










          • $begingroup$
            I see. Thanks alot.
            $endgroup$
            – user45765
            Dec 27 '18 at 1:22











          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%2f3053475%2frearrangement-of-summations-in-series-expansion-of-log-zetas%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          The point is that
          $$ left| sum_{n = 2}^N b_n - sum_{n = 2}^N a_n right| = left| pi (N) log left(1 - tfrac{1}{(N + 1)^s} right) right| leq CN^{1 - sigma}$$
          for some constant $C > 0$, and for sufficient large $N$.



          Since $1 - sigma < 0$, we have
          $$ lim_{N to infty} left| sum_{n = 2}^N b_n - sum_{n = 2}^N a_n right| = 0.$$



          So since $sum_{n = 2}^infty a_n$ converges, it must be true that $sum_{n = 2}^infty b_n$ converges too, and
          $ sum_{n = 2}^infty b_n = sum_{n = 2}^infty a_n$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Maybe this is a dumb question. Was this arrangement obvious at first sight? It is clear that it is natural to manipulate sum in that manner though it might not converge.
            $endgroup$
            – user45765
            Dec 27 '18 at 1:20










          • $begingroup$
            @user45765 I had to stare at it for a bit...
            $endgroup$
            – Kenny Wong
            Dec 27 '18 at 1:21










          • $begingroup$
            I see. Thanks alot.
            $endgroup$
            – user45765
            Dec 27 '18 at 1:22
















          1












          $begingroup$

          The point is that
          $$ left| sum_{n = 2}^N b_n - sum_{n = 2}^N a_n right| = left| pi (N) log left(1 - tfrac{1}{(N + 1)^s} right) right| leq CN^{1 - sigma}$$
          for some constant $C > 0$, and for sufficient large $N$.



          Since $1 - sigma < 0$, we have
          $$ lim_{N to infty} left| sum_{n = 2}^N b_n - sum_{n = 2}^N a_n right| = 0.$$



          So since $sum_{n = 2}^infty a_n$ converges, it must be true that $sum_{n = 2}^infty b_n$ converges too, and
          $ sum_{n = 2}^infty b_n = sum_{n = 2}^infty a_n$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Maybe this is a dumb question. Was this arrangement obvious at first sight? It is clear that it is natural to manipulate sum in that manner though it might not converge.
            $endgroup$
            – user45765
            Dec 27 '18 at 1:20










          • $begingroup$
            @user45765 I had to stare at it for a bit...
            $endgroup$
            – Kenny Wong
            Dec 27 '18 at 1:21










          • $begingroup$
            I see. Thanks alot.
            $endgroup$
            – user45765
            Dec 27 '18 at 1:22














          1












          1








          1





          $begingroup$

          The point is that
          $$ left| sum_{n = 2}^N b_n - sum_{n = 2}^N a_n right| = left| pi (N) log left(1 - tfrac{1}{(N + 1)^s} right) right| leq CN^{1 - sigma}$$
          for some constant $C > 0$, and for sufficient large $N$.



          Since $1 - sigma < 0$, we have
          $$ lim_{N to infty} left| sum_{n = 2}^N b_n - sum_{n = 2}^N a_n right| = 0.$$



          So since $sum_{n = 2}^infty a_n$ converges, it must be true that $sum_{n = 2}^infty b_n$ converges too, and
          $ sum_{n = 2}^infty b_n = sum_{n = 2}^infty a_n$.






          share|cite|improve this answer









          $endgroup$



          The point is that
          $$ left| sum_{n = 2}^N b_n - sum_{n = 2}^N a_n right| = left| pi (N) log left(1 - tfrac{1}{(N + 1)^s} right) right| leq CN^{1 - sigma}$$
          for some constant $C > 0$, and for sufficient large $N$.



          Since $1 - sigma < 0$, we have
          $$ lim_{N to infty} left| sum_{n = 2}^N b_n - sum_{n = 2}^N a_n right| = 0.$$



          So since $sum_{n = 2}^infty a_n$ converges, it must be true that $sum_{n = 2}^infty b_n$ converges too, and
          $ sum_{n = 2}^infty b_n = sum_{n = 2}^infty a_n$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 27 '18 at 1:14









          Kenny WongKenny Wong

          19.1k21441




          19.1k21441












          • $begingroup$
            Maybe this is a dumb question. Was this arrangement obvious at first sight? It is clear that it is natural to manipulate sum in that manner though it might not converge.
            $endgroup$
            – user45765
            Dec 27 '18 at 1:20










          • $begingroup$
            @user45765 I had to stare at it for a bit...
            $endgroup$
            – Kenny Wong
            Dec 27 '18 at 1:21










          • $begingroup$
            I see. Thanks alot.
            $endgroup$
            – user45765
            Dec 27 '18 at 1:22


















          • $begingroup$
            Maybe this is a dumb question. Was this arrangement obvious at first sight? It is clear that it is natural to manipulate sum in that manner though it might not converge.
            $endgroup$
            – user45765
            Dec 27 '18 at 1:20










          • $begingroup$
            @user45765 I had to stare at it for a bit...
            $endgroup$
            – Kenny Wong
            Dec 27 '18 at 1:21










          • $begingroup$
            I see. Thanks alot.
            $endgroup$
            – user45765
            Dec 27 '18 at 1:22
















          $begingroup$
          Maybe this is a dumb question. Was this arrangement obvious at first sight? It is clear that it is natural to manipulate sum in that manner though it might not converge.
          $endgroup$
          – user45765
          Dec 27 '18 at 1:20




          $begingroup$
          Maybe this is a dumb question. Was this arrangement obvious at first sight? It is clear that it is natural to manipulate sum in that manner though it might not converge.
          $endgroup$
          – user45765
          Dec 27 '18 at 1:20












          $begingroup$
          @user45765 I had to stare at it for a bit...
          $endgroup$
          – Kenny Wong
          Dec 27 '18 at 1:21




          $begingroup$
          @user45765 I had to stare at it for a bit...
          $endgroup$
          – Kenny Wong
          Dec 27 '18 at 1:21












          $begingroup$
          I see. Thanks alot.
          $endgroup$
          – user45765
          Dec 27 '18 at 1:22




          $begingroup$
          I see. Thanks alot.
          $endgroup$
          – user45765
          Dec 27 '18 at 1:22


















          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%2f3053475%2frearrangement-of-summations-in-series-expansion-of-log-zetas%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

          To store a contact into the json file from server.js file using a class in NodeJS

          Image
          0 I have created three files -contactPresentation.js, contactService.js, contacts.json. I will run the contactService.js file. And it will navigate to contactPresentation.js file that will only choose an option to either add or display a file. If add(), then add the data into add of service file.And the data will be added to the contacts.json file . And if display(), then display the display() of service file. And the display will take the data from contacts.json file. However, i know my code is flawed. Please correct it . I am also getting an error - this.contacts.push(new ContactService(userName, contactNumber, emailId)); ^ TypeError: Cannot read property 'push' of undefined at ContactService.add (e:NodeJSProjectcontactManager2contactService.js:57:23) at new C
          Read more

          Redirect URL with Chrome Remote Debugging Android Devices

          Image
          1 How can i redirect an URL with Chrome Remote Debugging Android Devices? like Requestly Chrome extension on desktop. google-chrome debugging url redirect testing share | improve this question edited Nov 24 '18 at 8:16 sachinjain024 14.7k 23 73 129 asked Nov 23 '18 at 16:09 Momen Momen 91 10
          Read more

          Dieringhausen

          Image
          Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
          Read more