What does $1+frac{1}{8}+frac{1}{27}+frac{1}{64}+frac{1}{125}+(frac{1}{n})^3$ equal to?












1












$begingroup$


I'm curious of what does this sum:



$1+frac{1}{8}+frac{1}{27}+frac{1}{64}+frac{1}{125}+frac{1}{216}+...+(frac{1}{n})^3$



or the Riemann zeta function: $zeta({3})$ approach. I watched a few 3Blue1Brown videos on YouTube, but it doesn't have any videos about what does that sum above approach. I don't know how to prove it, and I tried, but suddenly all of my tries lead to no results. Is this problem unsolved? If not, how would you prove it?






riemann-zeta







share|cite|improve this question











$endgroup$












  • $begingroup$
    It approaches, by definition, $zeta(3)ldots$ There's nothing to prove, except perhaps that the sum converges (obvious by integral test)
    $endgroup$
    – Brevan Ellefsen
    Nov 4 '18 at 22:30


















1












$begingroup$


I'm curious of what does this sum:



$1+frac{1}{8}+frac{1}{27}+frac{1}{64}+frac{1}{125}+frac{1}{216}+...+(frac{1}{n})^3$



or the Riemann zeta function: $zeta({3})$ approach. I watched a few 3Blue1Brown videos on YouTube, but it doesn't have any videos about what does that sum above approach. I don't know how to prove it, and I tried, but suddenly all of my tries lead to no results. Is this problem unsolved? If not, how would you prove it?






riemann-zeta







share|cite|improve this question











$endgroup$












  • $begingroup$
    It approaches, by definition, $zeta(3)ldots$ There's nothing to prove, except perhaps that the sum converges (obvious by integral test)
    $endgroup$
    – Brevan Ellefsen
    Nov 4 '18 at 22:30
















1












1








1





$begingroup$


I'm curious of what does this sum:



$1+frac{1}{8}+frac{1}{27}+frac{1}{64}+frac{1}{125}+frac{1}{216}+...+(frac{1}{n})^3$



or the Riemann zeta function: $zeta({3})$ approach. I watched a few 3Blue1Brown videos on YouTube, but it doesn't have any videos about what does that sum above approach. I don't know how to prove it, and I tried, but suddenly all of my tries lead to no results. Is this problem unsolved? If not, how would you prove it?






riemann-zeta







share|cite|improve this question











$endgroup$




I'm curious of what does this sum:



$1+frac{1}{8}+frac{1}{27}+frac{1}{64}+frac{1}{125}+frac{1}{216}+...+(frac{1}{n})^3$



or the Riemann zeta function: $zeta({3})$ approach. I watched a few 3Blue1Brown videos on YouTube, but it doesn't have any videos about what does that sum above approach. I don't know how to prove it, and I tried, but suddenly all of my tries lead to no results. Is this problem unsolved? If not, how would you prove it?






riemann-zeta




riemann-zeta






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 27 at 23:15







I-85a

















asked Nov 4 '18 at 22:27









I-85aI-85a

83




83












  • $begingroup$
    It approaches, by definition, $zeta(3)ldots$ There's nothing to prove, except perhaps that the sum converges (obvious by integral test)
    $endgroup$
    – Brevan Ellefsen
    Nov 4 '18 at 22:30




















  • $begingroup$
    It approaches, by definition, $zeta(3)ldots$ There's nothing to prove, except perhaps that the sum converges (obvious by integral test)
    $endgroup$
    – Brevan Ellefsen
    Nov 4 '18 at 22:30


















$begingroup$
It approaches, by definition, $zeta(3)ldots$ There's nothing to prove, except perhaps that the sum converges (obvious by integral test)
$endgroup$
– Brevan Ellefsen
Nov 4 '18 at 22:30






$begingroup$
It approaches, by definition, $zeta(3)ldots$ There's nothing to prove, except perhaps that the sum converges (obvious by integral test)
$endgroup$
– Brevan Ellefsen
Nov 4 '18 at 22:30












2 Answers
2






active

oldest

votes


















2












$begingroup$

$zeta(3)$ is a constant known as Apery's constant. Its value is approximately $1.2021ldots$ but as far as I know an exact value isn't known.



Some further reading:



https://en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant
http://mathworld.wolfram.com/AperysConstant.html






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I don't know what it is you think "isn't quite known". The "exact value" is $zeta(3)$. Apéry proved it is irrational.
    $endgroup$
    – Robert Israel
    Nov 4 '18 at 22:49










  • $begingroup$
    "The exact value of $zeta(3)$ is $zeta(3)$" isn't exactly an enlightening answer. I mean something in terms of - at least more known - constants, i.e. how we know $zeta(2) = pi^2/6$. A "closed form" value, perhaps? Whatever the term is.
    $endgroup$
    – Eevee Trainer
    Nov 4 '18 at 22:51










  • $begingroup$
    It is not likely that $zeta(3)$ has a closed form in terms of "better-known" constants, though of course this has not been proven. We don't even have a proof that it is transcendental.
    $endgroup$
    – Robert Israel
    Nov 4 '18 at 23:00










  • $begingroup$
    Well, there is $zeta(3) = -Psi''(1)/2$, but I don't know if you'd count that as "better-known".
    $endgroup$
    – Robert Israel
    Nov 5 '18 at 2:26










  • $begingroup$
    "The exact value of ζ(3) is ζ(3)" isn't exactly an enlightening answer." Why not? Why would not you say "the exact value of $pi$ isn't quite known"?
    $endgroup$
    – user
    Nov 5 '18 at 22:07





















2












$begingroup$

Well, $sum_{ngeq 1}frac{1}{n^3}$ is a number, precisely the value of the Riemann $zeta$ function at $s=3$. Since $xleq text{arctanh}(x)$ is a pretty tight approximation for $xto 0^+$,



$$zeta(3) = 1+sum_{ngeq 2}frac{1}{n^3} leq 1+frac{1}{2}sum_{ngeq 2}logleft(frac{n^3+1}{n^3-1}right)$$
but $prod_{ngeq 2}frac{n^3+1}{n^3-1}=prod_{ngeq 2}frac{n+1}{n-1}cdotfrac{n^2-n+1}{n^2+n+1}$ is a telescopic product convergent to $frac{3}{2}$, hence
$$ zeta(3) approx 1+frac{log 3-log 2}{2}.$$
By creative telescoping we also have the acceleration formula
$$ zeta(3)=sum_{ngeq 1}frac{1}{n^3}=frac{5}{2}sum_{ngeq 1}frac{(-1)^{n+1}}{n^3binom{2n}{n}}$$
(see my notes for a proof) which allowed Apery to prove that $zeta(3)notinmathbb{Q}$.

The irrationality of $zeta(5),zeta(7),zeta(9),ldots$ still is an open problem, like the conjecture $zeta(3)inpi^3mathbb{Q}$, which looks numerically very unlikely. On the other hand $sum_{ngeq 0}frac{(-1)^n}{(2n+1)^3}inpi^3mathbb{Q}$, as shown here.






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









    2












    $begingroup$

    $zeta(3)$ is a constant known as Apery's constant. Its value is approximately $1.2021ldots$ but as far as I know an exact value isn't known.



    Some further reading:



    https://en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant
    http://mathworld.wolfram.com/AperysConstant.html






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I don't know what it is you think "isn't quite known". The "exact value" is $zeta(3)$. Apéry proved it is irrational.
      $endgroup$
      – Robert Israel
      Nov 4 '18 at 22:49










    • $begingroup$
      "The exact value of $zeta(3)$ is $zeta(3)$" isn't exactly an enlightening answer. I mean something in terms of - at least more known - constants, i.e. how we know $zeta(2) = pi^2/6$. A "closed form" value, perhaps? Whatever the term is.
      $endgroup$
      – Eevee Trainer
      Nov 4 '18 at 22:51










    • $begingroup$
      It is not likely that $zeta(3)$ has a closed form in terms of "better-known" constants, though of course this has not been proven. We don't even have a proof that it is transcendental.
      $endgroup$
      – Robert Israel
      Nov 4 '18 at 23:00










    • $begingroup$
      Well, there is $zeta(3) = -Psi''(1)/2$, but I don't know if you'd count that as "better-known".
      $endgroup$
      – Robert Israel
      Nov 5 '18 at 2:26










    • $begingroup$
      "The exact value of ζ(3) is ζ(3)" isn't exactly an enlightening answer." Why not? Why would not you say "the exact value of $pi$ isn't quite known"?
      $endgroup$
      – user
      Nov 5 '18 at 22:07


















    2












    $begingroup$

    $zeta(3)$ is a constant known as Apery's constant. Its value is approximately $1.2021ldots$ but as far as I know an exact value isn't known.



    Some further reading:



    https://en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant
    http://mathworld.wolfram.com/AperysConstant.html






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I don't know what it is you think "isn't quite known". The "exact value" is $zeta(3)$. Apéry proved it is irrational.
      $endgroup$
      – Robert Israel
      Nov 4 '18 at 22:49










    • $begingroup$
      "The exact value of $zeta(3)$ is $zeta(3)$" isn't exactly an enlightening answer. I mean something in terms of - at least more known - constants, i.e. how we know $zeta(2) = pi^2/6$. A "closed form" value, perhaps? Whatever the term is.
      $endgroup$
      – Eevee Trainer
      Nov 4 '18 at 22:51










    • $begingroup$
      It is not likely that $zeta(3)$ has a closed form in terms of "better-known" constants, though of course this has not been proven. We don't even have a proof that it is transcendental.
      $endgroup$
      – Robert Israel
      Nov 4 '18 at 23:00










    • $begingroup$
      Well, there is $zeta(3) = -Psi''(1)/2$, but I don't know if you'd count that as "better-known".
      $endgroup$
      – Robert Israel
      Nov 5 '18 at 2:26










    • $begingroup$
      "The exact value of ζ(3) is ζ(3)" isn't exactly an enlightening answer." Why not? Why would not you say "the exact value of $pi$ isn't quite known"?
      $endgroup$
      – user
      Nov 5 '18 at 22:07
















    2












    2








    2





    $begingroup$

    $zeta(3)$ is a constant known as Apery's constant. Its value is approximately $1.2021ldots$ but as far as I know an exact value isn't known.



    Some further reading:



    https://en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant
    http://mathworld.wolfram.com/AperysConstant.html






    share|cite|improve this answer











    $endgroup$



    $zeta(3)$ is a constant known as Apery's constant. Its value is approximately $1.2021ldots$ but as far as I know an exact value isn't known.



    Some further reading:



    https://en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant
    http://mathworld.wolfram.com/AperysConstant.html







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 13 '18 at 9:02

























    answered Nov 4 '18 at 22:29









    Eevee TrainerEevee Trainer

    6,80311237




    6,80311237












    • $begingroup$
      I don't know what it is you think "isn't quite known". The "exact value" is $zeta(3)$. Apéry proved it is irrational.
      $endgroup$
      – Robert Israel
      Nov 4 '18 at 22:49










    • $begingroup$
      "The exact value of $zeta(3)$ is $zeta(3)$" isn't exactly an enlightening answer. I mean something in terms of - at least more known - constants, i.e. how we know $zeta(2) = pi^2/6$. A "closed form" value, perhaps? Whatever the term is.
      $endgroup$
      – Eevee Trainer
      Nov 4 '18 at 22:51










    • $begingroup$
      It is not likely that $zeta(3)$ has a closed form in terms of "better-known" constants, though of course this has not been proven. We don't even have a proof that it is transcendental.
      $endgroup$
      – Robert Israel
      Nov 4 '18 at 23:00










    • $begingroup$
      Well, there is $zeta(3) = -Psi''(1)/2$, but I don't know if you'd count that as "better-known".
      $endgroup$
      – Robert Israel
      Nov 5 '18 at 2:26










    • $begingroup$
      "The exact value of ζ(3) is ζ(3)" isn't exactly an enlightening answer." Why not? Why would not you say "the exact value of $pi$ isn't quite known"?
      $endgroup$
      – user
      Nov 5 '18 at 22:07




















    • $begingroup$
      I don't know what it is you think "isn't quite known". The "exact value" is $zeta(3)$. Apéry proved it is irrational.
      $endgroup$
      – Robert Israel
      Nov 4 '18 at 22:49










    • $begingroup$
      "The exact value of $zeta(3)$ is $zeta(3)$" isn't exactly an enlightening answer. I mean something in terms of - at least more known - constants, i.e. how we know $zeta(2) = pi^2/6$. A "closed form" value, perhaps? Whatever the term is.
      $endgroup$
      – Eevee Trainer
      Nov 4 '18 at 22:51










    • $begingroup$
      It is not likely that $zeta(3)$ has a closed form in terms of "better-known" constants, though of course this has not been proven. We don't even have a proof that it is transcendental.
      $endgroup$
      – Robert Israel
      Nov 4 '18 at 23:00










    • $begingroup$
      Well, there is $zeta(3) = -Psi''(1)/2$, but I don't know if you'd count that as "better-known".
      $endgroup$
      – Robert Israel
      Nov 5 '18 at 2:26










    • $begingroup$
      "The exact value of ζ(3) is ζ(3)" isn't exactly an enlightening answer." Why not? Why would not you say "the exact value of $pi$ isn't quite known"?
      $endgroup$
      – user
      Nov 5 '18 at 22:07


















    $begingroup$
    I don't know what it is you think "isn't quite known". The "exact value" is $zeta(3)$. Apéry proved it is irrational.
    $endgroup$
    – Robert Israel
    Nov 4 '18 at 22:49




    $begingroup$
    I don't know what it is you think "isn't quite known". The "exact value" is $zeta(3)$. Apéry proved it is irrational.
    $endgroup$
    – Robert Israel
    Nov 4 '18 at 22:49












    $begingroup$
    "The exact value of $zeta(3)$ is $zeta(3)$" isn't exactly an enlightening answer. I mean something in terms of - at least more known - constants, i.e. how we know $zeta(2) = pi^2/6$. A "closed form" value, perhaps? Whatever the term is.
    $endgroup$
    – Eevee Trainer
    Nov 4 '18 at 22:51




    $begingroup$
    "The exact value of $zeta(3)$ is $zeta(3)$" isn't exactly an enlightening answer. I mean something in terms of - at least more known - constants, i.e. how we know $zeta(2) = pi^2/6$. A "closed form" value, perhaps? Whatever the term is.
    $endgroup$
    – Eevee Trainer
    Nov 4 '18 at 22:51












    $begingroup$
    It is not likely that $zeta(3)$ has a closed form in terms of "better-known" constants, though of course this has not been proven. We don't even have a proof that it is transcendental.
    $endgroup$
    – Robert Israel
    Nov 4 '18 at 23:00




    $begingroup$
    It is not likely that $zeta(3)$ has a closed form in terms of "better-known" constants, though of course this has not been proven. We don't even have a proof that it is transcendental.
    $endgroup$
    – Robert Israel
    Nov 4 '18 at 23:00












    $begingroup$
    Well, there is $zeta(3) = -Psi''(1)/2$, but I don't know if you'd count that as "better-known".
    $endgroup$
    – Robert Israel
    Nov 5 '18 at 2:26




    $begingroup$
    Well, there is $zeta(3) = -Psi''(1)/2$, but I don't know if you'd count that as "better-known".
    $endgroup$
    – Robert Israel
    Nov 5 '18 at 2:26












    $begingroup$
    "The exact value of ζ(3) is ζ(3)" isn't exactly an enlightening answer." Why not? Why would not you say "the exact value of $pi$ isn't quite known"?
    $endgroup$
    – user
    Nov 5 '18 at 22:07






    $begingroup$
    "The exact value of ζ(3) is ζ(3)" isn't exactly an enlightening answer." Why not? Why would not you say "the exact value of $pi$ isn't quite known"?
    $endgroup$
    – user
    Nov 5 '18 at 22:07













    2












    $begingroup$

    Well, $sum_{ngeq 1}frac{1}{n^3}$ is a number, precisely the value of the Riemann $zeta$ function at $s=3$. Since $xleq text{arctanh}(x)$ is a pretty tight approximation for $xto 0^+$,



    $$zeta(3) = 1+sum_{ngeq 2}frac{1}{n^3} leq 1+frac{1}{2}sum_{ngeq 2}logleft(frac{n^3+1}{n^3-1}right)$$
    but $prod_{ngeq 2}frac{n^3+1}{n^3-1}=prod_{ngeq 2}frac{n+1}{n-1}cdotfrac{n^2-n+1}{n^2+n+1}$ is a telescopic product convergent to $frac{3}{2}$, hence
    $$ zeta(3) approx 1+frac{log 3-log 2}{2}.$$
    By creative telescoping we also have the acceleration formula
    $$ zeta(3)=sum_{ngeq 1}frac{1}{n^3}=frac{5}{2}sum_{ngeq 1}frac{(-1)^{n+1}}{n^3binom{2n}{n}}$$
    (see my notes for a proof) which allowed Apery to prove that $zeta(3)notinmathbb{Q}$.

    The irrationality of $zeta(5),zeta(7),zeta(9),ldots$ still is an open problem, like the conjecture $zeta(3)inpi^3mathbb{Q}$, which looks numerically very unlikely. On the other hand $sum_{ngeq 0}frac{(-1)^n}{(2n+1)^3}inpi^3mathbb{Q}$, as shown here.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Well, $sum_{ngeq 1}frac{1}{n^3}$ is a number, precisely the value of the Riemann $zeta$ function at $s=3$. Since $xleq text{arctanh}(x)$ is a pretty tight approximation for $xto 0^+$,



      $$zeta(3) = 1+sum_{ngeq 2}frac{1}{n^3} leq 1+frac{1}{2}sum_{ngeq 2}logleft(frac{n^3+1}{n^3-1}right)$$
      but $prod_{ngeq 2}frac{n^3+1}{n^3-1}=prod_{ngeq 2}frac{n+1}{n-1}cdotfrac{n^2-n+1}{n^2+n+1}$ is a telescopic product convergent to $frac{3}{2}$, hence
      $$ zeta(3) approx 1+frac{log 3-log 2}{2}.$$
      By creative telescoping we also have the acceleration formula
      $$ zeta(3)=sum_{ngeq 1}frac{1}{n^3}=frac{5}{2}sum_{ngeq 1}frac{(-1)^{n+1}}{n^3binom{2n}{n}}$$
      (see my notes for a proof) which allowed Apery to prove that $zeta(3)notinmathbb{Q}$.

      The irrationality of $zeta(5),zeta(7),zeta(9),ldots$ still is an open problem, like the conjecture $zeta(3)inpi^3mathbb{Q}$, which looks numerically very unlikely. On the other hand $sum_{ngeq 0}frac{(-1)^n}{(2n+1)^3}inpi^3mathbb{Q}$, as shown here.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Well, $sum_{ngeq 1}frac{1}{n^3}$ is a number, precisely the value of the Riemann $zeta$ function at $s=3$. Since $xleq text{arctanh}(x)$ is a pretty tight approximation for $xto 0^+$,



        $$zeta(3) = 1+sum_{ngeq 2}frac{1}{n^3} leq 1+frac{1}{2}sum_{ngeq 2}logleft(frac{n^3+1}{n^3-1}right)$$
        but $prod_{ngeq 2}frac{n^3+1}{n^3-1}=prod_{ngeq 2}frac{n+1}{n-1}cdotfrac{n^2-n+1}{n^2+n+1}$ is a telescopic product convergent to $frac{3}{2}$, hence
        $$ zeta(3) approx 1+frac{log 3-log 2}{2}.$$
        By creative telescoping we also have the acceleration formula
        $$ zeta(3)=sum_{ngeq 1}frac{1}{n^3}=frac{5}{2}sum_{ngeq 1}frac{(-1)^{n+1}}{n^3binom{2n}{n}}$$
        (see my notes for a proof) which allowed Apery to prove that $zeta(3)notinmathbb{Q}$.

        The irrationality of $zeta(5),zeta(7),zeta(9),ldots$ still is an open problem, like the conjecture $zeta(3)inpi^3mathbb{Q}$, which looks numerically very unlikely. On the other hand $sum_{ngeq 0}frac{(-1)^n}{(2n+1)^3}inpi^3mathbb{Q}$, as shown here.






        share|cite|improve this answer









        $endgroup$



        Well, $sum_{ngeq 1}frac{1}{n^3}$ is a number, precisely the value of the Riemann $zeta$ function at $s=3$. Since $xleq text{arctanh}(x)$ is a pretty tight approximation for $xto 0^+$,



        $$zeta(3) = 1+sum_{ngeq 2}frac{1}{n^3} leq 1+frac{1}{2}sum_{ngeq 2}logleft(frac{n^3+1}{n^3-1}right)$$
        but $prod_{ngeq 2}frac{n^3+1}{n^3-1}=prod_{ngeq 2}frac{n+1}{n-1}cdotfrac{n^2-n+1}{n^2+n+1}$ is a telescopic product convergent to $frac{3}{2}$, hence
        $$ zeta(3) approx 1+frac{log 3-log 2}{2}.$$
        By creative telescoping we also have the acceleration formula
        $$ zeta(3)=sum_{ngeq 1}frac{1}{n^3}=frac{5}{2}sum_{ngeq 1}frac{(-1)^{n+1}}{n^3binom{2n}{n}}$$
        (see my notes for a proof) which allowed Apery to prove that $zeta(3)notinmathbb{Q}$.

        The irrationality of $zeta(5),zeta(7),zeta(9),ldots$ still is an open problem, like the conjecture $zeta(3)inpi^3mathbb{Q}$, which looks numerically very unlikely. On the other hand $sum_{ngeq 0}frac{(-1)^n}{(2n+1)^3}inpi^3mathbb{Q}$, as shown here.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 4 '18 at 23:31









        Jack D'AurizioJack D'Aurizio

        290k33282664




        290k33282664






























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