Stirling type formula for Sum on $ln(n)^2$

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Is there a similar formula like the Stirling one on the sum over $ln(n)$ (take logarithms on its factorial representation):



$$sum_n=1^N ln(n) = N ln(N)-N+ln(N)/2+ln(2pi)/2+mathcalO(ln(N)/N)$$



but on the sum over its squares?



$$sum_n=1^N (ln(n))^2$$



I already advanced on getting good approximation on asymptotics integrating $ln(n)^2$ and arrive to correct terms till $mathcalO(N)$ order. But further advance is becoming hard for me in $mathcalO(ln(N))$ terms.



I am specially interested in $mathcalO(1)$ term.




sequences-and-series logarithms asymptotics




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  • 1




    Have you tried the Euler-Maclaurin formula?
    – Lord Shark the Unknown
    Aug 27 at 16:24










  • Maple says it's $$(ln(N)^2 - 2 ln(N)+2)N + fracln(N)^22 -frac ln left( 2,pi right)^22-fracpi^224 +fracgamma^22+gamma left( 1 right) + ldots $$
    – Robert Israel
    Aug 27 at 16:38











  • Sorry Robert, not sure what $$gamma(1)$$ means
    – 24th_moonshine
    Aug 27 at 16:46











  • Thanks @LordSharktheUnknown to your suggestion I could arrive to the O(ln(N)^2) term. But still no clear idea on how to get constant term. I saw some proof of Stirling constant term but no general method to apply here. Any idea on that?
    – 24th_moonshine
    Aug 27 at 16:48







  • 1




    $$sum _n=1^N log ^2(n)=sum _n=1^N left(undersetxto 0textlimfracpartial ^2n^xpartial x^2right)=undersetxto 0textlimfracpartial ^2partial x^2left(sum _n=1^N n^xright)=undersetxto 0textlimfracpartial ^2H_N^(-x)partial x^2=undersetxto 0textlim(textHarmonicNumber^(0,2)(N,-x))=fracgamma ^22-fracpi ^224-frac12 log ^2(2 pi )+gamma _1-zeta ^(2,0)(0,1+N)$$
    – Mariusz Iwaniuk
    Aug 29 at 14:33















up vote
2
down vote

favorite
2












Is there a similar formula like the Stirling one on the sum over $ln(n)$ (take logarithms on its factorial representation):



$$sum_n=1^N ln(n) = N ln(N)-N+ln(N)/2+ln(2pi)/2+mathcalO(ln(N)/N)$$



but on the sum over its squares?



$$sum_n=1^N (ln(n))^2$$



I already advanced on getting good approximation on asymptotics integrating $ln(n)^2$ and arrive to correct terms till $mathcalO(N)$ order. But further advance is becoming hard for me in $mathcalO(ln(N))$ terms.



I am specially interested in $mathcalO(1)$ term.




sequences-and-series logarithms asymptotics




share|cite|improve this question


















  • 1




    Have you tried the Euler-Maclaurin formula?
    – Lord Shark the Unknown
    Aug 27 at 16:24










  • Maple says it's $$(ln(N)^2 - 2 ln(N)+2)N + fracln(N)^22 -frac ln left( 2,pi right)^22-fracpi^224 +fracgamma^22+gamma left( 1 right) + ldots $$
    – Robert Israel
    Aug 27 at 16:38











  • Sorry Robert, not sure what $$gamma(1)$$ means
    – 24th_moonshine
    Aug 27 at 16:46











  • Thanks @LordSharktheUnknown to your suggestion I could arrive to the O(ln(N)^2) term. But still no clear idea on how to get constant term. I saw some proof of Stirling constant term but no general method to apply here. Any idea on that?
    – 24th_moonshine
    Aug 27 at 16:48







  • 1




    $$sum _n=1^N log ^2(n)=sum _n=1^N left(undersetxto 0textlimfracpartial ^2n^xpartial x^2right)=undersetxto 0textlimfracpartial ^2partial x^2left(sum _n=1^N n^xright)=undersetxto 0textlimfracpartial ^2H_N^(-x)partial x^2=undersetxto 0textlim(textHarmonicNumber^(0,2)(N,-x))=fracgamma ^22-fracpi ^224-frac12 log ^2(2 pi )+gamma _1-zeta ^(2,0)(0,1+N)$$
    – Mariusz Iwaniuk
    Aug 29 at 14:33













up vote
2
down vote

favorite
2









up vote
2
down vote

favorite
2






2





Is there a similar formula like the Stirling one on the sum over $ln(n)$ (take logarithms on its factorial representation):



$$sum_n=1^N ln(n) = N ln(N)-N+ln(N)/2+ln(2pi)/2+mathcalO(ln(N)/N)$$



but on the sum over its squares?



$$sum_n=1^N (ln(n))^2$$



I already advanced on getting good approximation on asymptotics integrating $ln(n)^2$ and arrive to correct terms till $mathcalO(N)$ order. But further advance is becoming hard for me in $mathcalO(ln(N))$ terms.



I am specially interested in $mathcalO(1)$ term.




sequences-and-series logarithms asymptotics




share|cite|improve this question














Is there a similar formula like the Stirling one on the sum over $ln(n)$ (take logarithms on its factorial representation):



$$sum_n=1^N ln(n) = N ln(N)-N+ln(N)/2+ln(2pi)/2+mathcalO(ln(N)/N)$$



but on the sum over its squares?



$$sum_n=1^N (ln(n))^2$$



I already advanced on getting good approximation on asymptotics integrating $ln(n)^2$ and arrive to correct terms till $mathcalO(N)$ order. But further advance is becoming hard for me in $mathcalO(ln(N))$ terms.



I am specially interested in $mathcalO(1)$ term.





sequences-and-series logarithms asymptotics





share|cite|improve this question













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edited Aug 27 at 17:34









ComplexYetTrivial

3,047626




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asked Aug 27 at 16:23









24th_moonshine

865




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  • 1




    Have you tried the Euler-Maclaurin formula?
    – Lord Shark the Unknown
    Aug 27 at 16:24










  • Maple says it's $$(ln(N)^2 - 2 ln(N)+2)N + fracln(N)^22 -frac ln left( 2,pi right)^22-fracpi^224 +fracgamma^22+gamma left( 1 right) + ldots $$
    – Robert Israel
    Aug 27 at 16:38











  • Sorry Robert, not sure what $$gamma(1)$$ means
    – 24th_moonshine
    Aug 27 at 16:46











  • Thanks @LordSharktheUnknown to your suggestion I could arrive to the O(ln(N)^2) term. But still no clear idea on how to get constant term. I saw some proof of Stirling constant term but no general method to apply here. Any idea on that?
    – 24th_moonshine
    Aug 27 at 16:48







  • 1




    $$sum _n=1^N log ^2(n)=sum _n=1^N left(undersetxto 0textlimfracpartial ^2n^xpartial x^2right)=undersetxto 0textlimfracpartial ^2partial x^2left(sum _n=1^N n^xright)=undersetxto 0textlimfracpartial ^2H_N^(-x)partial x^2=undersetxto 0textlim(textHarmonicNumber^(0,2)(N,-x))=fracgamma ^22-fracpi ^224-frac12 log ^2(2 pi )+gamma _1-zeta ^(2,0)(0,1+N)$$
    – Mariusz Iwaniuk
    Aug 29 at 14:33













  • 1




    Have you tried the Euler-Maclaurin formula?
    – Lord Shark the Unknown
    Aug 27 at 16:24










  • Maple says it's $$(ln(N)^2 - 2 ln(N)+2)N + fracln(N)^22 -frac ln left( 2,pi right)^22-fracpi^224 +fracgamma^22+gamma left( 1 right) + ldots $$
    – Robert Israel
    Aug 27 at 16:38











  • Sorry Robert, not sure what $$gamma(1)$$ means
    – 24th_moonshine
    Aug 27 at 16:46











  • Thanks @LordSharktheUnknown to your suggestion I could arrive to the O(ln(N)^2) term. But still no clear idea on how to get constant term. I saw some proof of Stirling constant term but no general method to apply here. Any idea on that?
    – 24th_moonshine
    Aug 27 at 16:48







  • 1




    $$sum _n=1^N log ^2(n)=sum _n=1^N left(undersetxto 0textlimfracpartial ^2n^xpartial x^2right)=undersetxto 0textlimfracpartial ^2partial x^2left(sum _n=1^N n^xright)=undersetxto 0textlimfracpartial ^2H_N^(-x)partial x^2=undersetxto 0textlim(textHarmonicNumber^(0,2)(N,-x))=fracgamma ^22-fracpi ^224-frac12 log ^2(2 pi )+gamma _1-zeta ^(2,0)(0,1+N)$$
    – Mariusz Iwaniuk
    Aug 29 at 14:33








1




1




Have you tried the Euler-Maclaurin formula?
– Lord Shark the Unknown
Aug 27 at 16:24




Have you tried the Euler-Maclaurin formula?
– Lord Shark the Unknown
Aug 27 at 16:24












Maple says it's $$(ln(N)^2 - 2 ln(N)+2)N + fracln(N)^22 -frac ln left( 2,pi right)^22-fracpi^224 +fracgamma^22+gamma left( 1 right) + ldots $$
– Robert Israel
Aug 27 at 16:38





Maple says it's $$(ln(N)^2 - 2 ln(N)+2)N + fracln(N)^22 -frac ln left( 2,pi right)^22-fracpi^224 +fracgamma^22+gamma left( 1 right) + ldots $$
– Robert Israel
Aug 27 at 16:38













Sorry Robert, not sure what $$gamma(1)$$ means
– 24th_moonshine
Aug 27 at 16:46





Sorry Robert, not sure what $$gamma(1)$$ means
– 24th_moonshine
Aug 27 at 16:46













Thanks @LordSharktheUnknown to your suggestion I could arrive to the O(ln(N)^2) term. But still no clear idea on how to get constant term. I saw some proof of Stirling constant term but no general method to apply here. Any idea on that?
– 24th_moonshine
Aug 27 at 16:48





Thanks @LordSharktheUnknown to your suggestion I could arrive to the O(ln(N)^2) term. But still no clear idea on how to get constant term. I saw some proof of Stirling constant term but no general method to apply here. Any idea on that?
– 24th_moonshine
Aug 27 at 16:48





1




1




$$sum _n=1^N log ^2(n)=sum _n=1^N left(undersetxto 0textlimfracpartial ^2n^xpartial x^2right)=undersetxto 0textlimfracpartial ^2partial x^2left(sum _n=1^N n^xright)=undersetxto 0textlimfracpartial ^2H_N^(-x)partial x^2=undersetxto 0textlim(textHarmonicNumber^(0,2)(N,-x))=fracgamma ^22-fracpi ^224-frac12 log ^2(2 pi )+gamma _1-zeta ^(2,0)(0,1+N)$$
– Mariusz Iwaniuk
Aug 29 at 14:33





$$sum _n=1^N log ^2(n)=sum _n=1^N left(undersetxto 0textlimfracpartial ^2n^xpartial x^2right)=undersetxto 0textlimfracpartial ^2partial x^2left(sum _n=1^N n^xright)=undersetxto 0textlimfracpartial ^2H_N^(-x)partial x^2=undersetxto 0textlim(textHarmonicNumber^(0,2)(N,-x))=fracgamma ^22-fracpi ^224-frac12 log ^2(2 pi )+gamma _1-zeta ^(2,0)(0,1+N)$$
– Mariusz Iwaniuk
Aug 29 at 14:33











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I've encountered an exercise recently which gives an asymptotic formula
$$
sum_1^N log^2(n)= left( n +frac 12right)log^2(n) -2n log(n) +2n + C + r_n quad [r_n to 0, C texta constant].
$$
Hope this could help.



The derivation TBA...






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    I've encountered an exercise recently which gives an asymptotic formula
    $$
    sum_1^N log^2(n)= left( n +frac 12right)log^2(n) -2n log(n) +2n + C + r_n quad [r_n to 0, C texta constant].
    $$
    Hope this could help.



    The derivation TBA...






    share|cite|improve this answer
























      up vote
      0
      down vote













      I've encountered an exercise recently which gives an asymptotic formula
      $$
      sum_1^N log^2(n)= left( n +frac 12right)log^2(n) -2n log(n) +2n + C + r_n quad [r_n to 0, C texta constant].
      $$
      Hope this could help.



      The derivation TBA...






      share|cite|improve this answer






















        up vote
        0
        down vote










        up vote
        0
        down vote









        I've encountered an exercise recently which gives an asymptotic formula
        $$
        sum_1^N log^2(n)= left( n +frac 12right)log^2(n) -2n log(n) +2n + C + r_n quad [r_n to 0, C texta constant].
        $$
        Hope this could help.



        The derivation TBA...






        share|cite|improve this answer












        I've encountered an exercise recently which gives an asymptotic formula
        $$
        sum_1^N log^2(n)= left( n +frac 12right)log^2(n) -2n log(n) +2n + C + r_n quad [r_n to 0, C texta constant].
        $$
        Hope this could help.



        The derivation TBA...







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 27 at 16:42









        xbh

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