limit of ratio of partition function












0












$begingroup$


Does the following limit exists?



$$lim_{n rightarrow infty} frac{p(n)}{p(n-5)}$$



where $p(n)$ denote the partition function.



If this limit exists, is it equal to 1?



Kindly share your thoughts.



Thank you.






integer-partitions







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Does the following limit exists?



    $$lim_{n rightarrow infty} frac{p(n)}{p(n-5)}$$



    where $p(n)$ denote the partition function.



    If this limit exists, is it equal to 1?



    Kindly share your thoughts.



    Thank you.






    integer-partitions







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Does the following limit exists?



      $$lim_{n rightarrow infty} frac{p(n)}{p(n-5)}$$



      where $p(n)$ denote the partition function.



      If this limit exists, is it equal to 1?



      Kindly share your thoughts.



      Thank you.






      integer-partitions







      share|cite|improve this question









      $endgroup$




      Does the following limit exists?



      $$lim_{n rightarrow infty} frac{p(n)}{p(n-5)}$$



      where $p(n)$ denote the partition function.



      If this limit exists, is it equal to 1?



      Kindly share your thoughts.



      Thank you.






      integer-partitions




      integer-partitions






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 3 at 6:40









      GA316GA316

      2,6591232




      2,6591232






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Using $$p(n) sim frac{e^{pisqrt{frac{2n}3}}}{4sqrt3,n} $$ compute $frac{p(n)}{p(n-a)}$, take logarithms and expand as series for large values og $n$. This would give
          $$log left(frac{p(n)}{p(n-a)}right)=frac{pi a}{sqrt{6n}}-frac{a}{2
          n}+Oleft(frac{1}{n^{3/2}}right)$$           and continuing with Taylor series using $x=e^{log(x)}$, then
          $$frac{p(n)}{p(n-a)}=1+frac{pi a }{sqrt{6n}}+frac{a left(pi ^2 a-6right)}{12
          n}+Oleft(frac{1}{n^{3/2}}right)$$           which shows the limit (already given in answers) but also a quite good approximation of the ratio.



          Using $n=50$ and $a=5$, the exact value is $frac{102113}{44567}approx 2.291$ while the above expansion would give $frac{114+20 sqrt{3} pi +5 pi ^2}{120} approx 2.268$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you. Very insightful.
            $endgroup$
            – GA316
            Jan 3 at 8:30



















          1












          $begingroup$

          $p(n) sim frac1{4sqrt3n} e^{pisqrt{2n/3}}$ according to OEIS and GTM177 Chapter II.

          Hence, $$lim_{n rightarrow infty} frac{p(n)}{p(n-5)}=lim_{n rightarrow infty} frac{frac1{4sqrt3n} e^{pisqrt{2n/3}}}{frac1{4sqrt3(n-5)} e^{pisqrt{2(n-5)/3}}}=1$$






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            Using $$p(n) sim frac{e^{pisqrt{frac{2n}3}}}{4sqrt3,n} $$ compute $frac{p(n)}{p(n-a)}$, take logarithms and expand as series for large values og $n$. This would give
            $$log left(frac{p(n)}{p(n-a)}right)=frac{pi a}{sqrt{6n}}-frac{a}{2
            n}+Oleft(frac{1}{n^{3/2}}right)$$             and continuing with Taylor series using $x=e^{log(x)}$, then
            $$frac{p(n)}{p(n-a)}=1+frac{pi a }{sqrt{6n}}+frac{a left(pi ^2 a-6right)}{12
            n}+Oleft(frac{1}{n^{3/2}}right)$$             which shows the limit (already given in answers) but also a quite good approximation of the ratio.



            Using $n=50$ and $a=5$, the exact value is $frac{102113}{44567}approx 2.291$ while the above expansion would give $frac{114+20 sqrt{3} pi +5 pi ^2}{120} approx 2.268$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you. Very insightful.
              $endgroup$
              – GA316
              Jan 3 at 8:30
















            1












            $begingroup$

            Using $$p(n) sim frac{e^{pisqrt{frac{2n}3}}}{4sqrt3,n} $$ compute $frac{p(n)}{p(n-a)}$, take logarithms and expand as series for large values og $n$. This would give
            $$log left(frac{p(n)}{p(n-a)}right)=frac{pi a}{sqrt{6n}}-frac{a}{2
            n}+Oleft(frac{1}{n^{3/2}}right)$$             and continuing with Taylor series using $x=e^{log(x)}$, then
            $$frac{p(n)}{p(n-a)}=1+frac{pi a }{sqrt{6n}}+frac{a left(pi ^2 a-6right)}{12
            n}+Oleft(frac{1}{n^{3/2}}right)$$             which shows the limit (already given in answers) but also a quite good approximation of the ratio.



            Using $n=50$ and $a=5$, the exact value is $frac{102113}{44567}approx 2.291$ while the above expansion would give $frac{114+20 sqrt{3} pi +5 pi ^2}{120} approx 2.268$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you. Very insightful.
              $endgroup$
              – GA316
              Jan 3 at 8:30














            1












            1








            1





            $begingroup$

            Using $$p(n) sim frac{e^{pisqrt{frac{2n}3}}}{4sqrt3,n} $$ compute $frac{p(n)}{p(n-a)}$, take logarithms and expand as series for large values og $n$. This would give
            $$log left(frac{p(n)}{p(n-a)}right)=frac{pi a}{sqrt{6n}}-frac{a}{2
            n}+Oleft(frac{1}{n^{3/2}}right)$$             and continuing with Taylor series using $x=e^{log(x)}$, then
            $$frac{p(n)}{p(n-a)}=1+frac{pi a }{sqrt{6n}}+frac{a left(pi ^2 a-6right)}{12
            n}+Oleft(frac{1}{n^{3/2}}right)$$             which shows the limit (already given in answers) but also a quite good approximation of the ratio.



            Using $n=50$ and $a=5$, the exact value is $frac{102113}{44567}approx 2.291$ while the above expansion would give $frac{114+20 sqrt{3} pi +5 pi ^2}{120} approx 2.268$






            share|cite|improve this answer









            $endgroup$



            Using $$p(n) sim frac{e^{pisqrt{frac{2n}3}}}{4sqrt3,n} $$ compute $frac{p(n)}{p(n-a)}$, take logarithms and expand as series for large values og $n$. This would give
            $$log left(frac{p(n)}{p(n-a)}right)=frac{pi a}{sqrt{6n}}-frac{a}{2
            n}+Oleft(frac{1}{n^{3/2}}right)$$             and continuing with Taylor series using $x=e^{log(x)}$, then
            $$frac{p(n)}{p(n-a)}=1+frac{pi a }{sqrt{6n}}+frac{a left(pi ^2 a-6right)}{12
            n}+Oleft(frac{1}{n^{3/2}}right)$$             which shows the limit (already given in answers) but also a quite good approximation of the ratio.



            Using $n=50$ and $a=5$, the exact value is $frac{102113}{44567}approx 2.291$ while the above expansion would give $frac{114+20 sqrt{3} pi +5 pi ^2}{120} approx 2.268$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 3 at 8:01









            Claude LeiboviciClaude Leibovici

            120k1157132




            120k1157132












            • $begingroup$
              Thank you. Very insightful.
              $endgroup$
              – GA316
              Jan 3 at 8:30


















            • $begingroup$
              Thank you. Very insightful.
              $endgroup$
              – GA316
              Jan 3 at 8:30
















            $begingroup$
            Thank you. Very insightful.
            $endgroup$
            – GA316
            Jan 3 at 8:30




            $begingroup$
            Thank you. Very insightful.
            $endgroup$
            – GA316
            Jan 3 at 8:30











            1












            $begingroup$

            $p(n) sim frac1{4sqrt3n} e^{pisqrt{2n/3}}$ according to OEIS and GTM177 Chapter II.

            Hence, $$lim_{n rightarrow infty} frac{p(n)}{p(n-5)}=lim_{n rightarrow infty} frac{frac1{4sqrt3n} e^{pisqrt{2n/3}}}{frac1{4sqrt3(n-5)} e^{pisqrt{2(n-5)/3}}}=1$$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              $p(n) sim frac1{4sqrt3n} e^{pisqrt{2n/3}}$ according to OEIS and GTM177 Chapter II.

              Hence, $$lim_{n rightarrow infty} frac{p(n)}{p(n-5)}=lim_{n rightarrow infty} frac{frac1{4sqrt3n} e^{pisqrt{2n/3}}}{frac1{4sqrt3(n-5)} e^{pisqrt{2(n-5)/3}}}=1$$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                $p(n) sim frac1{4sqrt3n} e^{pisqrt{2n/3}}$ according to OEIS and GTM177 Chapter II.

                Hence, $$lim_{n rightarrow infty} frac{p(n)}{p(n-5)}=lim_{n rightarrow infty} frac{frac1{4sqrt3n} e^{pisqrt{2n/3}}}{frac1{4sqrt3(n-5)} e^{pisqrt{2(n-5)/3}}}=1$$






                share|cite|improve this answer









                $endgroup$



                $p(n) sim frac1{4sqrt3n} e^{pisqrt{2n/3}}$ according to OEIS and GTM177 Chapter II.

                Hence, $$lim_{n rightarrow infty} frac{p(n)}{p(n-5)}=lim_{n rightarrow infty} frac{frac1{4sqrt3n} e^{pisqrt{2n/3}}}{frac1{4sqrt3(n-5)} e^{pisqrt{2(n-5)/3}}}=1$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 3 at 7:18









                Kemono ChenKemono Chen

                3,0521743




                3,0521743






























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