A closed form for infinite series $ sum _1 ^infty frac {1}{n^3} = zeta (3) ?$












1












$begingroup$


It is well known that $$ sum _1 ^infty frac {1}{n^2} = frac {pi ^2}{6}$$ and $$ sum _1 ^infty frac {1}{n^4} = frac {pi ^4}{90}$$



We also know that $$ sum _1 ^infty frac {1}{n^3} $$



$$=1.202056903159594285399738161511449990764986292340498881.....$$



My question is:



Do we have a closed form for this series besides $$ sum _1 ^infty frac {1}{n^3} = zeta (3) ?$$






functions zeta-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant gives several fast converging infinite sums
    $endgroup$
    – Henry
    Apr 8 '18 at 10:05












  • $begingroup$
    Thanks, I meant something like $$pi^? /?$$
    $endgroup$
    – Mohammad Riazi-Kermani
    Apr 8 '18 at 10:08










  • $begingroup$
    Last time I checked, no, there is no closed form for $zeta(n)$ where $n in 2mathbb{N}+1$.
    $endgroup$
    – Rócherz
    Apr 8 '18 at 10:18








  • 1




    $begingroup$
    @MohammadRiazi-Kermani Quoting from Establishing zeta(3) as a definite integral and its computation on MO: "it is conjectured (but not known) that the values $zeta(2n+1)$ are ... algebraically independent of one another, and of $pi$.".
    $endgroup$
    – dxiv
    Apr 9 '18 at 3:33
















1












$begingroup$


It is well known that $$ sum _1 ^infty frac {1}{n^2} = frac {pi ^2}{6}$$ and $$ sum _1 ^infty frac {1}{n^4} = frac {pi ^4}{90}$$



We also know that $$ sum _1 ^infty frac {1}{n^3} $$



$$=1.202056903159594285399738161511449990764986292340498881.....$$



My question is:



Do we have a closed form for this series besides $$ sum _1 ^infty frac {1}{n^3} = zeta (3) ?$$






functions zeta-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant gives several fast converging infinite sums
    $endgroup$
    – Henry
    Apr 8 '18 at 10:05












  • $begingroup$
    Thanks, I meant something like $$pi^? /?$$
    $endgroup$
    – Mohammad Riazi-Kermani
    Apr 8 '18 at 10:08










  • $begingroup$
    Last time I checked, no, there is no closed form for $zeta(n)$ where $n in 2mathbb{N}+1$.
    $endgroup$
    – Rócherz
    Apr 8 '18 at 10:18








  • 1




    $begingroup$
    @MohammadRiazi-Kermani Quoting from Establishing zeta(3) as a definite integral and its computation on MO: "it is conjectured (but not known) that the values $zeta(2n+1)$ are ... algebraically independent of one another, and of $pi$.".
    $endgroup$
    – dxiv
    Apr 9 '18 at 3:33














1












1








1





$begingroup$


It is well known that $$ sum _1 ^infty frac {1}{n^2} = frac {pi ^2}{6}$$ and $$ sum _1 ^infty frac {1}{n^4} = frac {pi ^4}{90}$$



We also know that $$ sum _1 ^infty frac {1}{n^3} $$



$$=1.202056903159594285399738161511449990764986292340498881.....$$



My question is:



Do we have a closed form for this series besides $$ sum _1 ^infty frac {1}{n^3} = zeta (3) ?$$






functions zeta-functions







share|cite|improve this question











$endgroup$




It is well known that $$ sum _1 ^infty frac {1}{n^2} = frac {pi ^2}{6}$$ and $$ sum _1 ^infty frac {1}{n^4} = frac {pi ^4}{90}$$



We also know that $$ sum _1 ^infty frac {1}{n^3} $$



$$=1.202056903159594285399738161511449990764986292340498881.....$$



My question is:



Do we have a closed form for this series besides $$ sum _1 ^infty frac {1}{n^3} = zeta (3) ?$$






functions zeta-functions




functions zeta-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 9 '18 at 6:30







Mohammad Riazi-Kermani

















asked Apr 8 '18 at 10:03









Mohammad Riazi-KermaniMohammad Riazi-Kermani

41.6k42061




41.6k42061












  • $begingroup$
    en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant gives several fast converging infinite sums
    $endgroup$
    – Henry
    Apr 8 '18 at 10:05












  • $begingroup$
    Thanks, I meant something like $$pi^? /?$$
    $endgroup$
    – Mohammad Riazi-Kermani
    Apr 8 '18 at 10:08










  • $begingroup$
    Last time I checked, no, there is no closed form for $zeta(n)$ where $n in 2mathbb{N}+1$.
    $endgroup$
    – Rócherz
    Apr 8 '18 at 10:18








  • 1




    $begingroup$
    @MohammadRiazi-Kermani Quoting from Establishing zeta(3) as a definite integral and its computation on MO: "it is conjectured (but not known) that the values $zeta(2n+1)$ are ... algebraically independent of one another, and of $pi$.".
    $endgroup$
    – dxiv
    Apr 9 '18 at 3:33


















  • $begingroup$
    en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant gives several fast converging infinite sums
    $endgroup$
    – Henry
    Apr 8 '18 at 10:05












  • $begingroup$
    Thanks, I meant something like $$pi^? /?$$
    $endgroup$
    – Mohammad Riazi-Kermani
    Apr 8 '18 at 10:08










  • $begingroup$
    Last time I checked, no, there is no closed form for $zeta(n)$ where $n in 2mathbb{N}+1$.
    $endgroup$
    – Rócherz
    Apr 8 '18 at 10:18








  • 1




    $begingroup$
    @MohammadRiazi-Kermani Quoting from Establishing zeta(3) as a definite integral and its computation on MO: "it is conjectured (but not known) that the values $zeta(2n+1)$ are ... algebraically independent of one another, and of $pi$.".
    $endgroup$
    – dxiv
    Apr 9 '18 at 3:33
















$begingroup$
en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant gives several fast converging infinite sums
$endgroup$
– Henry
Apr 8 '18 at 10:05






$begingroup$
en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant gives several fast converging infinite sums
$endgroup$
– Henry
Apr 8 '18 at 10:05














$begingroup$
Thanks, I meant something like $$pi^? /?$$
$endgroup$
– Mohammad Riazi-Kermani
Apr 8 '18 at 10:08




$begingroup$
Thanks, I meant something like $$pi^? /?$$
$endgroup$
– Mohammad Riazi-Kermani
Apr 8 '18 at 10:08












$begingroup$
Last time I checked, no, there is no closed form for $zeta(n)$ where $n in 2mathbb{N}+1$.
$endgroup$
– Rócherz
Apr 8 '18 at 10:18






$begingroup$
Last time I checked, no, there is no closed form for $zeta(n)$ where $n in 2mathbb{N}+1$.
$endgroup$
– Rócherz
Apr 8 '18 at 10:18






1




1




$begingroup$
@MohammadRiazi-Kermani Quoting from Establishing zeta(3) as a definite integral and its computation on MO: "it is conjectured (but not known) that the values $zeta(2n+1)$ are ... algebraically independent of one another, and of $pi$.".
$endgroup$
– dxiv
Apr 9 '18 at 3:33




$begingroup$
@MohammadRiazi-Kermani Quoting from Establishing zeta(3) as a definite integral and its computation on MO: "it is conjectured (but not known) that the values $zeta(2n+1)$ are ... algebraically independent of one another, and of $pi$.".
$endgroup$
– dxiv
Apr 9 '18 at 3:33










2 Answers
2






active

oldest

votes


















4












$begingroup$

Apéry's Constant



As mentioned in comments
$$
sum_{n=1}^inftyfrac1{n^3}=zeta(3)tag1
$$
is also known as Apéry's Constant. There is no closed form in terms of a rational multiple of an integer power of $pi$.



We also have
$$
sum_{n=1}^inftyfrac1{(2n-1)^3}=frac78zeta(3)tag2
$$



However, as shown in this answer, if we alternate the series in $(2)$, we get
$$
sum_{n=1}^inftyfrac{(-1)^{n-1}}{(2n-1)^3}=frac{pi^3}{32}tag3
$$
which is also known as $beta(3)$, the Dirichlet beta function.




Computation of $boldsymbol{zeta(3)}$



The series in $(1)$ converges very slowly. To get the sum to about $10$ places, we would need to sum $100000$ terms.



If we use the Euler-Maclaurin Sum Formula, we get that
$$
sum_{k=1}^nfrac1{k^3}=smallzeta(3)-frac1{2n^2}+frac1{2n^3}-frac1{4n^4}+frac1{12n^6}-frac1{12n^8}+frac3{20n^{10}}-frac5{12n^{12}}+O!left(frac1{n^{14}}right)tag4
$$
Using $n=10$ in $(4)$, we get
$$
sum_{k=1}^inftyfrac1{k^3}=1.2020569031596tag5
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the information. Equation (3) in particular may be of further interest.
    $endgroup$
    – Mohammad Riazi-Kermani
    Apr 9 '18 at 11:04





















2












$begingroup$

In addition to robojohn's answer, there are some formulas expressing $zeta(3)$ (and other odd zeta values) in terms of powers of $pi$, but these are not clsoed forms. The most well known ones are due to Plouffe and Borwein & Bradley:



$$
begin{aligned}
zeta(3)&=frac{7pi^3}{180}-2sum_{n=1}^infty frac{1}{n^3(e^{2pi n}-1)},\
sum_{n=1}^infty frac{1}{n^3,binom {2n}n} &= -frac{4}{3},zeta(3)+frac{pisqrt{3}}{2cdot 3^2},left(zeta(2, tfrac{1}{3})-zeta(2,tfrac{2}{3}) right).
end{aligned}
$$



Moreover, in this Math.SE post we have:



$$
frac{3}{2},zeta(3) = frac{pi^3}{24}sqrt{2}-2sum_{k=1}^infty frac{1}{k^3(e^{pi ksqrt{2}}-1)}-sum_{k=1}^inftyfrac{1}{k^3(e^{2pi ksqrt{2}}-1)}.
$$



You can also check out this paper by Vepstas, which provides a nice generalization to some of these identities.






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









    4












    $begingroup$

    Apéry's Constant



    As mentioned in comments
    $$
    sum_{n=1}^inftyfrac1{n^3}=zeta(3)tag1
    $$
    is also known as Apéry's Constant. There is no closed form in terms of a rational multiple of an integer power of $pi$.



    We also have
    $$
    sum_{n=1}^inftyfrac1{(2n-1)^3}=frac78zeta(3)tag2
    $$



    However, as shown in this answer, if we alternate the series in $(2)$, we get
    $$
    sum_{n=1}^inftyfrac{(-1)^{n-1}}{(2n-1)^3}=frac{pi^3}{32}tag3
    $$
    which is also known as $beta(3)$, the Dirichlet beta function.




    Computation of $boldsymbol{zeta(3)}$



    The series in $(1)$ converges very slowly. To get the sum to about $10$ places, we would need to sum $100000$ terms.



    If we use the Euler-Maclaurin Sum Formula, we get that
    $$
    sum_{k=1}^nfrac1{k^3}=smallzeta(3)-frac1{2n^2}+frac1{2n^3}-frac1{4n^4}+frac1{12n^6}-frac1{12n^8}+frac3{20n^{10}}-frac5{12n^{12}}+O!left(frac1{n^{14}}right)tag4
    $$
    Using $n=10$ in $(4)$, we get
    $$
    sum_{k=1}^inftyfrac1{k^3}=1.2020569031596tag5
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks for the information. Equation (3) in particular may be of further interest.
      $endgroup$
      – Mohammad Riazi-Kermani
      Apr 9 '18 at 11:04


















    4












    $begingroup$

    Apéry's Constant



    As mentioned in comments
    $$
    sum_{n=1}^inftyfrac1{n^3}=zeta(3)tag1
    $$
    is also known as Apéry's Constant. There is no closed form in terms of a rational multiple of an integer power of $pi$.



    We also have
    $$
    sum_{n=1}^inftyfrac1{(2n-1)^3}=frac78zeta(3)tag2
    $$



    However, as shown in this answer, if we alternate the series in $(2)$, we get
    $$
    sum_{n=1}^inftyfrac{(-1)^{n-1}}{(2n-1)^3}=frac{pi^3}{32}tag3
    $$
    which is also known as $beta(3)$, the Dirichlet beta function.




    Computation of $boldsymbol{zeta(3)}$



    The series in $(1)$ converges very slowly. To get the sum to about $10$ places, we would need to sum $100000$ terms.



    If we use the Euler-Maclaurin Sum Formula, we get that
    $$
    sum_{k=1}^nfrac1{k^3}=smallzeta(3)-frac1{2n^2}+frac1{2n^3}-frac1{4n^4}+frac1{12n^6}-frac1{12n^8}+frac3{20n^{10}}-frac5{12n^{12}}+O!left(frac1{n^{14}}right)tag4
    $$
    Using $n=10$ in $(4)$, we get
    $$
    sum_{k=1}^inftyfrac1{k^3}=1.2020569031596tag5
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks for the information. Equation (3) in particular may be of further interest.
      $endgroup$
      – Mohammad Riazi-Kermani
      Apr 9 '18 at 11:04
















    4












    4








    4





    $begingroup$

    Apéry's Constant



    As mentioned in comments
    $$
    sum_{n=1}^inftyfrac1{n^3}=zeta(3)tag1
    $$
    is also known as Apéry's Constant. There is no closed form in terms of a rational multiple of an integer power of $pi$.



    We also have
    $$
    sum_{n=1}^inftyfrac1{(2n-1)^3}=frac78zeta(3)tag2
    $$



    However, as shown in this answer, if we alternate the series in $(2)$, we get
    $$
    sum_{n=1}^inftyfrac{(-1)^{n-1}}{(2n-1)^3}=frac{pi^3}{32}tag3
    $$
    which is also known as $beta(3)$, the Dirichlet beta function.




    Computation of $boldsymbol{zeta(3)}$



    The series in $(1)$ converges very slowly. To get the sum to about $10$ places, we would need to sum $100000$ terms.



    If we use the Euler-Maclaurin Sum Formula, we get that
    $$
    sum_{k=1}^nfrac1{k^3}=smallzeta(3)-frac1{2n^2}+frac1{2n^3}-frac1{4n^4}+frac1{12n^6}-frac1{12n^8}+frac3{20n^{10}}-frac5{12n^{12}}+O!left(frac1{n^{14}}right)tag4
    $$
    Using $n=10$ in $(4)$, we get
    $$
    sum_{k=1}^inftyfrac1{k^3}=1.2020569031596tag5
    $$






    share|cite|improve this answer











    $endgroup$



    Apéry's Constant



    As mentioned in comments
    $$
    sum_{n=1}^inftyfrac1{n^3}=zeta(3)tag1
    $$
    is also known as Apéry's Constant. There is no closed form in terms of a rational multiple of an integer power of $pi$.



    We also have
    $$
    sum_{n=1}^inftyfrac1{(2n-1)^3}=frac78zeta(3)tag2
    $$



    However, as shown in this answer, if we alternate the series in $(2)$, we get
    $$
    sum_{n=1}^inftyfrac{(-1)^{n-1}}{(2n-1)^3}=frac{pi^3}{32}tag3
    $$
    which is also known as $beta(3)$, the Dirichlet beta function.




    Computation of $boldsymbol{zeta(3)}$



    The series in $(1)$ converges very slowly. To get the sum to about $10$ places, we would need to sum $100000$ terms.



    If we use the Euler-Maclaurin Sum Formula, we get that
    $$
    sum_{k=1}^nfrac1{k^3}=smallzeta(3)-frac1{2n^2}+frac1{2n^3}-frac1{4n^4}+frac1{12n^6}-frac1{12n^8}+frac3{20n^{10}}-frac5{12n^{12}}+O!left(frac1{n^{14}}right)tag4
    $$
    Using $n=10$ in $(4)$, we get
    $$
    sum_{k=1}^inftyfrac1{k^3}=1.2020569031596tag5
    $$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Apr 9 '18 at 7:48

























    answered Apr 9 '18 at 7:08









    robjohn♦robjohn

    267k27308632




    267k27308632












    • $begingroup$
      Thanks for the information. Equation (3) in particular may be of further interest.
      $endgroup$
      – Mohammad Riazi-Kermani
      Apr 9 '18 at 11:04




















    • $begingroup$
      Thanks for the information. Equation (3) in particular may be of further interest.
      $endgroup$
      – Mohammad Riazi-Kermani
      Apr 9 '18 at 11:04


















    $begingroup$
    Thanks for the information. Equation (3) in particular may be of further interest.
    $endgroup$
    – Mohammad Riazi-Kermani
    Apr 9 '18 at 11:04






    $begingroup$
    Thanks for the information. Equation (3) in particular may be of further interest.
    $endgroup$
    – Mohammad Riazi-Kermani
    Apr 9 '18 at 11:04













    2












    $begingroup$

    In addition to robojohn's answer, there are some formulas expressing $zeta(3)$ (and other odd zeta values) in terms of powers of $pi$, but these are not clsoed forms. The most well known ones are due to Plouffe and Borwein & Bradley:



    $$
    begin{aligned}
    zeta(3)&=frac{7pi^3}{180}-2sum_{n=1}^infty frac{1}{n^3(e^{2pi n}-1)},\
    sum_{n=1}^infty frac{1}{n^3,binom {2n}n} &= -frac{4}{3},zeta(3)+frac{pisqrt{3}}{2cdot 3^2},left(zeta(2, tfrac{1}{3})-zeta(2,tfrac{2}{3}) right).
    end{aligned}
    $$



    Moreover, in this Math.SE post we have:



    $$
    frac{3}{2},zeta(3) = frac{pi^3}{24}sqrt{2}-2sum_{k=1}^infty frac{1}{k^3(e^{pi ksqrt{2}}-1)}-sum_{k=1}^inftyfrac{1}{k^3(e^{2pi ksqrt{2}}-1)}.
    $$



    You can also check out this paper by Vepstas, which provides a nice generalization to some of these identities.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      In addition to robojohn's answer, there are some formulas expressing $zeta(3)$ (and other odd zeta values) in terms of powers of $pi$, but these are not clsoed forms. The most well known ones are due to Plouffe and Borwein & Bradley:



      $$
      begin{aligned}
      zeta(3)&=frac{7pi^3}{180}-2sum_{n=1}^infty frac{1}{n^3(e^{2pi n}-1)},\
      sum_{n=1}^infty frac{1}{n^3,binom {2n}n} &= -frac{4}{3},zeta(3)+frac{pisqrt{3}}{2cdot 3^2},left(zeta(2, tfrac{1}{3})-zeta(2,tfrac{2}{3}) right).
      end{aligned}
      $$



      Moreover, in this Math.SE post we have:



      $$
      frac{3}{2},zeta(3) = frac{pi^3}{24}sqrt{2}-2sum_{k=1}^infty frac{1}{k^3(e^{pi ksqrt{2}}-1)}-sum_{k=1}^inftyfrac{1}{k^3(e^{2pi ksqrt{2}}-1)}.
      $$



      You can also check out this paper by Vepstas, which provides a nice generalization to some of these identities.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        In addition to robojohn's answer, there are some formulas expressing $zeta(3)$ (and other odd zeta values) in terms of powers of $pi$, but these are not clsoed forms. The most well known ones are due to Plouffe and Borwein & Bradley:



        $$
        begin{aligned}
        zeta(3)&=frac{7pi^3}{180}-2sum_{n=1}^infty frac{1}{n^3(e^{2pi n}-1)},\
        sum_{n=1}^infty frac{1}{n^3,binom {2n}n} &= -frac{4}{3},zeta(3)+frac{pisqrt{3}}{2cdot 3^2},left(zeta(2, tfrac{1}{3})-zeta(2,tfrac{2}{3}) right).
        end{aligned}
        $$



        Moreover, in this Math.SE post we have:



        $$
        frac{3}{2},zeta(3) = frac{pi^3}{24}sqrt{2}-2sum_{k=1}^infty frac{1}{k^3(e^{pi ksqrt{2}}-1)}-sum_{k=1}^inftyfrac{1}{k^3(e^{2pi ksqrt{2}}-1)}.
        $$



        You can also check out this paper by Vepstas, which provides a nice generalization to some of these identities.






        share|cite|improve this answer









        $endgroup$



        In addition to robojohn's answer, there are some formulas expressing $zeta(3)$ (and other odd zeta values) in terms of powers of $pi$, but these are not clsoed forms. The most well known ones are due to Plouffe and Borwein & Bradley:



        $$
        begin{aligned}
        zeta(3)&=frac{7pi^3}{180}-2sum_{n=1}^infty frac{1}{n^3(e^{2pi n}-1)},\
        sum_{n=1}^infty frac{1}{n^3,binom {2n}n} &= -frac{4}{3},zeta(3)+frac{pisqrt{3}}{2cdot 3^2},left(zeta(2, tfrac{1}{3})-zeta(2,tfrac{2}{3}) right).
        end{aligned}
        $$



        Moreover, in this Math.SE post we have:



        $$
        frac{3}{2},zeta(3) = frac{pi^3}{24}sqrt{2}-2sum_{k=1}^infty frac{1}{k^3(e^{pi ksqrt{2}}-1)}-sum_{k=1}^inftyfrac{1}{k^3(e^{2pi ksqrt{2}}-1)}.
        $$



        You can also check out this paper by Vepstas, which provides a nice generalization to some of these identities.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 13 '18 at 9:33









        KlangenKlangen

        1,72811334




        1,72811334






























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