Why is it so hard to find a closed form for $sum_{n=1}^infty frac{1}{n^3}$?












3












$begingroup$


The series $sum_{n=1}^infty frac{1}{n^p}$ converges for $p>1$; I have known this result since I took calculus in my freshman year. It is also known that $$sum_{n=1}^infty frac{1}{n^2}=frac{pi^2}{6} text{ and } sum_{n=1}^infty frac{1}{n^4}=frac{pi^4}{90}$$



I learned a few years later--while taking a history and philosophy of mathematics course of all places--that the precise value for which the series converges for $p=3$ is still unknown. Doing some brief investigation--using Wolfram|Alpha and Wikipedia--it appears that the result is defined in terms of the Riemann zeta function, i.e. $$sum_{n=1}^infty frac{1}{n^3}=zeta(3)approx 1.2020569...$$



I have been told that a closed form has not been found. My question is why the mechanisms we have developed that enable us to find a closed form in the previous two cases fail in the third case? What it is that makes the $p=3$ case much more difficult? Is there even a closed form for the $p=3$ case, and for that matter, for all odd-numbered cases? I have attempted to ask one of my math professors this question, but I did not quite understand the explanation at the time.



For added context, I am currently doing an undergraduate degree in statistics. As such, I have only taken two semesters of real analysis, and one semester of complex analysis. I also feel like this question might have been have been asked on this site before; if so, please point me in the right direction.










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  • $begingroup$
    Search "Apery" on this site. Your exact question was asked by someone. I will point to it if I find it.
    $endgroup$
    – Srini
    Dec 30 '16 at 3:52






  • 17




    $begingroup$
    The question is somewhat backwards: what is surprising is that one can find closed forms for the other series! There is no reason at all for a series, no matter how nice the general term, has a sum with a nice closed form.
    $endgroup$
    – Mariano Suárez-Álvarez
    Dec 30 '16 at 3:52










  • $begingroup$
    That one can solve the even cases can be traced back to the fact that when you plug $pi$ into the Taylor expansion of $sin(x)$ you should get 0. There is nothing similar for the odd cases.
    $endgroup$
    – Fabian
    Dec 30 '16 at 4:00










  • $begingroup$
    This thread may help you.
    $endgroup$
    – Rohan
    Dec 30 '16 at 4:11






  • 2




    $begingroup$
    See this similar question we can compute a closed-form of $zeta(2n)$ because it is the derivative of $cos(z)/sin(z)$ at $z=1/2$, but $zeta(2n+1)$ is the derivative of $Gamma'(z)/Gamma(z)$ at $z=1/2$ which is much more complicated
    $endgroup$
    – reuns
    Dec 30 '16 at 10:29


















3












$begingroup$


The series $sum_{n=1}^infty frac{1}{n^p}$ converges for $p>1$; I have known this result since I took calculus in my freshman year. It is also known that $$sum_{n=1}^infty frac{1}{n^2}=frac{pi^2}{6} text{ and } sum_{n=1}^infty frac{1}{n^4}=frac{pi^4}{90}$$



I learned a few years later--while taking a history and philosophy of mathematics course of all places--that the precise value for which the series converges for $p=3$ is still unknown. Doing some brief investigation--using Wolfram|Alpha and Wikipedia--it appears that the result is defined in terms of the Riemann zeta function, i.e. $$sum_{n=1}^infty frac{1}{n^3}=zeta(3)approx 1.2020569...$$



I have been told that a closed form has not been found. My question is why the mechanisms we have developed that enable us to find a closed form in the previous two cases fail in the third case? What it is that makes the $p=3$ case much more difficult? Is there even a closed form for the $p=3$ case, and for that matter, for all odd-numbered cases? I have attempted to ask one of my math professors this question, but I did not quite understand the explanation at the time.



For added context, I am currently doing an undergraduate degree in statistics. As such, I have only taken two semesters of real analysis, and one semester of complex analysis. I also feel like this question might have been have been asked on this site before; if so, please point me in the right direction.










share|cite|improve this question









$endgroup$












  • $begingroup$
    Search "Apery" on this site. Your exact question was asked by someone. I will point to it if I find it.
    $endgroup$
    – Srini
    Dec 30 '16 at 3:52






  • 17




    $begingroup$
    The question is somewhat backwards: what is surprising is that one can find closed forms for the other series! There is no reason at all for a series, no matter how nice the general term, has a sum with a nice closed form.
    $endgroup$
    – Mariano Suárez-Álvarez
    Dec 30 '16 at 3:52










  • $begingroup$
    That one can solve the even cases can be traced back to the fact that when you plug $pi$ into the Taylor expansion of $sin(x)$ you should get 0. There is nothing similar for the odd cases.
    $endgroup$
    – Fabian
    Dec 30 '16 at 4:00










  • $begingroup$
    This thread may help you.
    $endgroup$
    – Rohan
    Dec 30 '16 at 4:11






  • 2




    $begingroup$
    See this similar question we can compute a closed-form of $zeta(2n)$ because it is the derivative of $cos(z)/sin(z)$ at $z=1/2$, but $zeta(2n+1)$ is the derivative of $Gamma'(z)/Gamma(z)$ at $z=1/2$ which is much more complicated
    $endgroup$
    – reuns
    Dec 30 '16 at 10:29
















3












3








3


1



$begingroup$


The series $sum_{n=1}^infty frac{1}{n^p}$ converges for $p>1$; I have known this result since I took calculus in my freshman year. It is also known that $$sum_{n=1}^infty frac{1}{n^2}=frac{pi^2}{6} text{ and } sum_{n=1}^infty frac{1}{n^4}=frac{pi^4}{90}$$



I learned a few years later--while taking a history and philosophy of mathematics course of all places--that the precise value for which the series converges for $p=3$ is still unknown. Doing some brief investigation--using Wolfram|Alpha and Wikipedia--it appears that the result is defined in terms of the Riemann zeta function, i.e. $$sum_{n=1}^infty frac{1}{n^3}=zeta(3)approx 1.2020569...$$



I have been told that a closed form has not been found. My question is why the mechanisms we have developed that enable us to find a closed form in the previous two cases fail in the third case? What it is that makes the $p=3$ case much more difficult? Is there even a closed form for the $p=3$ case, and for that matter, for all odd-numbered cases? I have attempted to ask one of my math professors this question, but I did not quite understand the explanation at the time.



For added context, I am currently doing an undergraduate degree in statistics. As such, I have only taken two semesters of real analysis, and one semester of complex analysis. I also feel like this question might have been have been asked on this site before; if so, please point me in the right direction.










share|cite|improve this question









$endgroup$




The series $sum_{n=1}^infty frac{1}{n^p}$ converges for $p>1$; I have known this result since I took calculus in my freshman year. It is also known that $$sum_{n=1}^infty frac{1}{n^2}=frac{pi^2}{6} text{ and } sum_{n=1}^infty frac{1}{n^4}=frac{pi^4}{90}$$



I learned a few years later--while taking a history and philosophy of mathematics course of all places--that the precise value for which the series converges for $p=3$ is still unknown. Doing some brief investigation--using Wolfram|Alpha and Wikipedia--it appears that the result is defined in terms of the Riemann zeta function, i.e. $$sum_{n=1}^infty frac{1}{n^3}=zeta(3)approx 1.2020569...$$



I have been told that a closed form has not been found. My question is why the mechanisms we have developed that enable us to find a closed form in the previous two cases fail in the third case? What it is that makes the $p=3$ case much more difficult? Is there even a closed form for the $p=3$ case, and for that matter, for all odd-numbered cases? I have attempted to ask one of my math professors this question, but I did not quite understand the explanation at the time.



For added context, I am currently doing an undergraduate degree in statistics. As such, I have only taken two semesters of real analysis, and one semester of complex analysis. I also feel like this question might have been have been asked on this site before; if so, please point me in the right direction.







calculus sequences-and-series complex-analysis






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asked Dec 30 '16 at 3:45









Douglas FirDouglas Fir

1,224918




1,224918












  • $begingroup$
    Search "Apery" on this site. Your exact question was asked by someone. I will point to it if I find it.
    $endgroup$
    – Srini
    Dec 30 '16 at 3:52






  • 17




    $begingroup$
    The question is somewhat backwards: what is surprising is that one can find closed forms for the other series! There is no reason at all for a series, no matter how nice the general term, has a sum with a nice closed form.
    $endgroup$
    – Mariano Suárez-Álvarez
    Dec 30 '16 at 3:52










  • $begingroup$
    That one can solve the even cases can be traced back to the fact that when you plug $pi$ into the Taylor expansion of $sin(x)$ you should get 0. There is nothing similar for the odd cases.
    $endgroup$
    – Fabian
    Dec 30 '16 at 4:00










  • $begingroup$
    This thread may help you.
    $endgroup$
    – Rohan
    Dec 30 '16 at 4:11






  • 2




    $begingroup$
    See this similar question we can compute a closed-form of $zeta(2n)$ because it is the derivative of $cos(z)/sin(z)$ at $z=1/2$, but $zeta(2n+1)$ is the derivative of $Gamma'(z)/Gamma(z)$ at $z=1/2$ which is much more complicated
    $endgroup$
    – reuns
    Dec 30 '16 at 10:29




















  • $begingroup$
    Search "Apery" on this site. Your exact question was asked by someone. I will point to it if I find it.
    $endgroup$
    – Srini
    Dec 30 '16 at 3:52






  • 17




    $begingroup$
    The question is somewhat backwards: what is surprising is that one can find closed forms for the other series! There is no reason at all for a series, no matter how nice the general term, has a sum with a nice closed form.
    $endgroup$
    – Mariano Suárez-Álvarez
    Dec 30 '16 at 3:52










  • $begingroup$
    That one can solve the even cases can be traced back to the fact that when you plug $pi$ into the Taylor expansion of $sin(x)$ you should get 0. There is nothing similar for the odd cases.
    $endgroup$
    – Fabian
    Dec 30 '16 at 4:00










  • $begingroup$
    This thread may help you.
    $endgroup$
    – Rohan
    Dec 30 '16 at 4:11






  • 2




    $begingroup$
    See this similar question we can compute a closed-form of $zeta(2n)$ because it is the derivative of $cos(z)/sin(z)$ at $z=1/2$, but $zeta(2n+1)$ is the derivative of $Gamma'(z)/Gamma(z)$ at $z=1/2$ which is much more complicated
    $endgroup$
    – reuns
    Dec 30 '16 at 10:29


















$begingroup$
Search "Apery" on this site. Your exact question was asked by someone. I will point to it if I find it.
$endgroup$
– Srini
Dec 30 '16 at 3:52




$begingroup$
Search "Apery" on this site. Your exact question was asked by someone. I will point to it if I find it.
$endgroup$
– Srini
Dec 30 '16 at 3:52




17




17




$begingroup$
The question is somewhat backwards: what is surprising is that one can find closed forms for the other series! There is no reason at all for a series, no matter how nice the general term, has a sum with a nice closed form.
$endgroup$
– Mariano Suárez-Álvarez
Dec 30 '16 at 3:52




$begingroup$
The question is somewhat backwards: what is surprising is that one can find closed forms for the other series! There is no reason at all for a series, no matter how nice the general term, has a sum with a nice closed form.
$endgroup$
– Mariano Suárez-Álvarez
Dec 30 '16 at 3:52












$begingroup$
That one can solve the even cases can be traced back to the fact that when you plug $pi$ into the Taylor expansion of $sin(x)$ you should get 0. There is nothing similar for the odd cases.
$endgroup$
– Fabian
Dec 30 '16 at 4:00




$begingroup$
That one can solve the even cases can be traced back to the fact that when you plug $pi$ into the Taylor expansion of $sin(x)$ you should get 0. There is nothing similar for the odd cases.
$endgroup$
– Fabian
Dec 30 '16 at 4:00












$begingroup$
This thread may help you.
$endgroup$
– Rohan
Dec 30 '16 at 4:11




$begingroup$
This thread may help you.
$endgroup$
– Rohan
Dec 30 '16 at 4:11




2




2




$begingroup$
See this similar question we can compute a closed-form of $zeta(2n)$ because it is the derivative of $cos(z)/sin(z)$ at $z=1/2$, but $zeta(2n+1)$ is the derivative of $Gamma'(z)/Gamma(z)$ at $z=1/2$ which is much more complicated
$endgroup$
– reuns
Dec 30 '16 at 10:29






$begingroup$
See this similar question we can compute a closed-form of $zeta(2n)$ because it is the derivative of $cos(z)/sin(z)$ at $z=1/2$, but $zeta(2n+1)$ is the derivative of $Gamma'(z)/Gamma(z)$ at $z=1/2$ which is much more complicated
$endgroup$
– reuns
Dec 30 '16 at 10:29












2 Answers
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Roger Apéry proved in 1978 that $zeta(3)=displaystyle sum_{n=1}^{infty} n^{-3}$ is irrational, however the irrationality of $displaystylefrac{zeta(3)}{pi^3}$ is still an open problem. No closed form for its value is currently known.



There are however some formulas expressing $zeta(3)$ (and other odd zeta values) in terms of powers of $pi$, but these are not closed 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











$endgroup$





















    1












    $begingroup$

    It can be shown that $frac{sin pi x}{pi x}=prod_{kge 1}(1-frac{x^2}{k^2})$. The left-hand side's Maclaurin series can be used to write $pi^{2n}$ in terms of $zeta(2k),,1le kle n$. There is no similar theoretical treatment of $zeta(m)$ for odd $mge 3$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      An interesting attempt to extend Euler's method to $zeta(3)$ can be found here: math.stackexchange.com/questions/1459709/…
      $endgroup$
      – Klangen
      Dec 13 '18 at 9:48













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    2 Answers
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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes









    1












    $begingroup$

    Roger Apéry proved in 1978 that $zeta(3)=displaystyle sum_{n=1}^{infty} n^{-3}$ is irrational, however the irrationality of $displaystylefrac{zeta(3)}{pi^3}$ is still an open problem. No closed form for its value is currently known.



    There are however some formulas expressing $zeta(3)$ (and other odd zeta values) in terms of powers of $pi$, but these are not closed 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











    $endgroup$


















      1












      $begingroup$

      Roger Apéry proved in 1978 that $zeta(3)=displaystyle sum_{n=1}^{infty} n^{-3}$ is irrational, however the irrationality of $displaystylefrac{zeta(3)}{pi^3}$ is still an open problem. No closed form for its value is currently known.



      There are however some formulas expressing $zeta(3)$ (and other odd zeta values) in terms of powers of $pi$, but these are not closed 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











      $endgroup$
















        1












        1








        1





        $begingroup$

        Roger Apéry proved in 1978 that $zeta(3)=displaystyle sum_{n=1}^{infty} n^{-3}$ is irrational, however the irrationality of $displaystylefrac{zeta(3)}{pi^3}$ is still an open problem. No closed form for its value is currently known.



        There are however some formulas expressing $zeta(3)$ (and other odd zeta values) in terms of powers of $pi$, but these are not closed 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











        $endgroup$



        Roger Apéry proved in 1978 that $zeta(3)=displaystyle sum_{n=1}^{infty} n^{-3}$ is irrational, however the irrationality of $displaystylefrac{zeta(3)}{pi^3}$ is still an open problem. No closed form for its value is currently known.



        There are however some formulas expressing $zeta(3)$ (and other odd zeta values) in terms of powers of $pi$, but these are not closed 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








        edited Dec 13 '18 at 9:08

























        answered Dec 13 '18 at 8:58









        KlangenKlangen

        1,75811334




        1,75811334























            1












            $begingroup$

            It can be shown that $frac{sin pi x}{pi x}=prod_{kge 1}(1-frac{x^2}{k^2})$. The left-hand side's Maclaurin series can be used to write $pi^{2n}$ in terms of $zeta(2k),,1le kle n$. There is no similar theoretical treatment of $zeta(m)$ for odd $mge 3$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              An interesting attempt to extend Euler's method to $zeta(3)$ can be found here: math.stackexchange.com/questions/1459709/…
              $endgroup$
              – Klangen
              Dec 13 '18 at 9:48


















            1












            $begingroup$

            It can be shown that $frac{sin pi x}{pi x}=prod_{kge 1}(1-frac{x^2}{k^2})$. The left-hand side's Maclaurin series can be used to write $pi^{2n}$ in terms of $zeta(2k),,1le kle n$. There is no similar theoretical treatment of $zeta(m)$ for odd $mge 3$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              An interesting attempt to extend Euler's method to $zeta(3)$ can be found here: math.stackexchange.com/questions/1459709/…
              $endgroup$
              – Klangen
              Dec 13 '18 at 9:48
















            1












            1








            1





            $begingroup$

            It can be shown that $frac{sin pi x}{pi x}=prod_{kge 1}(1-frac{x^2}{k^2})$. The left-hand side's Maclaurin series can be used to write $pi^{2n}$ in terms of $zeta(2k),,1le kle n$. There is no similar theoretical treatment of $zeta(m)$ for odd $mge 3$.






            share|cite|improve this answer









            $endgroup$



            It can be shown that $frac{sin pi x}{pi x}=prod_{kge 1}(1-frac{x^2}{k^2})$. The left-hand side's Maclaurin series can be used to write $pi^{2n}$ in terms of $zeta(2k),,1le kle n$. There is no similar theoretical treatment of $zeta(m)$ for odd $mge 3$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 13 '18 at 9:18









            J.G.J.G.

            28k22844




            28k22844












            • $begingroup$
              An interesting attempt to extend Euler's method to $zeta(3)$ can be found here: math.stackexchange.com/questions/1459709/…
              $endgroup$
              – Klangen
              Dec 13 '18 at 9:48




















            • $begingroup$
              An interesting attempt to extend Euler's method to $zeta(3)$ can be found here: math.stackexchange.com/questions/1459709/…
              $endgroup$
              – Klangen
              Dec 13 '18 at 9:48


















            $begingroup$
            An interesting attempt to extend Euler's method to $zeta(3)$ can be found here: math.stackexchange.com/questions/1459709/…
            $endgroup$
            – Klangen
            Dec 13 '18 at 9:48






            $begingroup$
            An interesting attempt to extend Euler's method to $zeta(3)$ can be found here: math.stackexchange.com/questions/1459709/…
            $endgroup$
            – Klangen
            Dec 13 '18 at 9:48




















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