What is an elementary method to compute $\zeta (3)$ which can be understood by a high school student? I know how to compute $\zeta (2)$ but not $\zeta (3)$.
Any help is very much appreciated.
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6What do you mean by "to compute $\zeta(3)$"? What would be the desirable form of the answer? – Start wearing purple May 08 '13 at 10:35
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By definition $$ \zeta(3)=\sum_{n=1}^\infty\frac1{n^3} $$ This can be used to compute $\zeta(3)$, but the convergence is fairly slow (you'd need to sum $200,000$ terms to get $10$ places).
There are ways to accelerate the convergence such as Euler's transform. The Euler-Maclaurin Sum Formula can also be used to accelerate the series. However, these acceleration methods are probably beyond most high school students (Euler's transform might be accessible).
robjohn
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2Also, whereas the value of $\zeta(2)$ has a nice closed form, the same cannot be said for $\zeta(3)$ in the sense that we do not know if such a nice closed form exists or not (although presumably it does not) – Andrew D May 08 '13 at 11:15
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2@AndrewD: It is true that Apéry's Constant is irrational, but it is not known if it is transcendental. Furthermore, $$ \sum_{k=0}^\infty\frac1{(2k+1)^3}=\frac78\zeta(3) $$ but $$ \sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^3}=\frac{\pi^3}{32} $$ – robjohn May 08 '13 at 11:38
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@robjohn Wow! Great comparison! I have seen the first one but never the second one. Can I ask where you found the second equation? – zerosofthezeta Dec 03 '13 at 20:47
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1@zeta: I wrote this paper which gives recurrence relations for $\zeta(2n)$ and $\beta(2n+1)$. – robjohn Dec 03 '13 at 20:57
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The dropbox site linked above is no longer accessible. This answer has the same information on the Dirichlet Beta function. – robjohn Feb 14 '19 at 19:34