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Apery gave,

$\begin{aligned} \zeta(3) &= \frac{5}{2}\,\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3\,\binom {2k}k}\end{aligned}$

J. Borwein and D. Bradley found this can be generalized to $\zeta(4n+3)$. Define the functions,

$\begin{aligned} &B(a_0)=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k\,^{a_0}\,\binom {2k}k}\\ &B(a_0,a_1,a_2,\dots)=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k\,^{a_0}\,\binom {2k}k}\; \sum_{p=1}^{k-1} \frac{1}{p\,^{a_1}}\;\sum_{q=1}^{k-1} \frac{1}{q\,^{a_2}}\;\dots \end{aligned}$

then,

$\begin{aligned} \frac{2}{5}\,\zeta(3) &= B(3)\\ \frac{2}{5}\,\zeta(7) &= B(7)+5B(3,4)\\ \frac{2}{5}\,\zeta(11) &= B(11)+5B(7,4)-\frac{15}{2}B(3,8)+\frac{25}{2}B(3,4,4)\\ \frac{2}{5}\,\zeta(15) &= B(15)+5B(11,4)-\frac{15}{2}B(7,8) +\frac{25}{2}B(7,4,4) +\frac{130}{2}B(3,12)-\frac{225}{6}B(3,8,4)+ \frac{125}{6}B(3,4,4,4) \end{aligned}$

My old version of Mathematica can verify the first three, but not the one for $\zeta(15)$. Can someone check if indeed the last is true, or if I misinterpreted the (different) Borwein-Bradley formulation given in page 2-3 of Apery-Like Formulae for $\zeta(4n+3)$?

share|improve this question
    
Maple evaluates the last equation to $0.4000122352945228 = -0.4309839386322169$, so I am assuming the equation is wrong. –  GEdgar May 27 '12 at 17:04
    
@GEdgar. Thanks. I must need new contact lenses, as the problem was that I mistyped one of the coefficients. I'll delete the question in a while. –  Tito Piezas III May 27 '12 at 17:24
    
Rather, "...I should have the question deleted". It should have been $\frac{130}{6}B(3,12)$. –  Tito Piezas III May 27 '12 at 17:38
    
That works.${}{}$ –  GEdgar May 27 '12 at 18:37
    
@TitoPiezasIII Seem to work fine. –  Balarka Sen Feb 8 at 20:18
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