Sorry I'm really out of any sort of number theory... is this a known product? \begin{align} P & = \prod_{n=2}^\infty \left(1-\dfrac1{n^p}\right) \end{align} As a special case I would be interested in $p=3$. Thank you very much
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1It can be written in terms of the gamma function. – Sangchul Lee Feb 18 '16 at 07:02
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that would work, can you tell me how? – Dac0 Feb 18 '16 at 07:03
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1The basic idea is to factor $(1-n^{-p})$ into linear terms and utilize the Weierstrass product for $\Gamma(z)$ to simply the resulting expression. For a technical reason it is easier to work with partial products first. – Sangchul Lee Feb 18 '16 at 07:08
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Of course the valuable fact is an understandable proof, which I hope see from an user to learn it. But here you have a tool for the future (type product (1-1/k^3), k=2 to infinity) in INFINITE PRODUCTS here https://www.wolframalpha.com/input/?i=product+(1-1%2Fk%5E3),+k%3D2+to+infinity – Feb 18 '16 at 07:25
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Just give the infinite product to maple and you will get a closed form! – Mhenni Benghorbal Feb 18 '16 at 07:26
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Three related questions. – Lucian Feb 18 '16 at 08:18
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From the Weierstrass product for $\Gamma(z)$, we can easily find that
$$ \frac{e^{(\gamma-1)z}}{\Gamma(2-z)} = \prod_{n=2}^{\infty} \left( 1 - \frac{z}{n} \right) e^{z/n}. $$
Writing
$$ \prod_{n=2}^{\infty} \left(1 - \frac{1}{n^p}\right) = \prod_{n=2}^{\infty} \prod_{\omega : \omega^p = 1} \left( 1 - \frac{\omega}{n} \right) e^{\omega/n} $$
and interchanging the order of summation, we have
$$ \prod_{n=2}^{\infty} \left(1 - \frac{1}{n^p}\right) = \prod_{\omega : \omega^p = 1} \frac{1}{\Gamma(2-\omega)}. $$
For some special choice of $p$ (such as $p = 2$ or $p = 3$), this can be simplified further.
Sangchul Lee
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