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












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$begingroup$


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







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$endgroup$

















    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
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          $begingroup$

          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











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          1 Answer
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          active

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          $begingroup$

          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


















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