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Can anyone compute $$\prod_{i=1}^{\infty}{1+(\frac{k}{i})^3}$$ for integer k? Can it be done in closed form, using only elementary functions, without the use of the Gamma function? For k=1, the closed expressions are known to include $\cosh(.)$ and $\pi$.

I hope to use this one day to find a closed expression of $\zeta(3)$. If we could compute $$\prod_{i=1}^{\infty}{1+(\frac{z}{i})^3}$$ for complex z, it follows that $\zeta(3)$ is the coefficient for $z^3$ in its expansion.

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Here is a closed form for the product

$$ {\frac {1}{{z}^{2}\Gamma \left( z+1 \right) \Gamma \left( -\frac{z}{2}-\frac{iz\sqrt {3}}{2} \right) \Gamma \left( -\frac{z}{2}+\frac{iz\sqrt {3}}{2} \right)}},\quad i=\sqrt{-1},$$

where $\Gamma(z)$ is the gamma function.

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    $\begingroup$ Of course that doesn't help @Cuc much, because a series expansion gives (you guessed it) $1 + \zeta(3) z^3 + \ldots$. $\endgroup$ – Robert Israel Feb 12 '14 at 8:44
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    $\begingroup$ How did @MhenniBenghorbal achieve the closed form ? $\endgroup$ – Zophikel Jan 29 '18 at 17:12
  • $\begingroup$ Where's the k in the expression? And can cosh be used to get rid of the Gamma function? $\endgroup$ – Cuc Aug 29 at 19:21
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Just as an addendum to Robert Israel's comment. The Taylor expansion of the formula given in Mhenni Benghorbal's answer is $$1+\zeta (3) z^3 + \left(\frac{\zeta (3)^2}{2}-\frac{\pi ^6}{1890}\right)z^6+(\frac{1}{6} \left(\zeta (3)^3+2 \zeta (9)\right)-\frac{\pi ^6 \zeta (3)}{1890})z^9 +O\left(z^{12}\right)$$ which I find quite nice.

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