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How find this $$\prod_{n=2}^{\infty}\left(1-\dfrac{1}{n^6}\right)$$

I think we can find this value have closed form $$\prod_{n=2}^{\infty}\left(1-\dfrac{1}{n^{2k}}\right)$$ since $$1-\dfrac{1}{n^6}=\left(1-\dfrac{1}{n^3}\right)\left(1+\dfrac{1}{n^3}\right)$$ so I think we must find this $$\prod_{n=2}^{\infty}\left(1-\dfrac{1}{n^3}\right)$$ and $$\prod_{n=2}^{\infty}\left(1+\dfrac{1}{n^3}\right)$$

Thank you

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What have you tried? You might consider finding a common denominator and factoring. Or you could try taking log of the product. – SpamIAm Feb 13 '14 at 4:26
A related problem. – Mhenni Benghorbal Feb 13 '14 at 5:20
up vote 3 down vote accepted

As Mhenni Benghorbal mentioned, there is a similar problem which has been treated yesterday and I shall use a similar approach to the one he proposed.

$$\prod_{n=2}^{m}\left(1-\dfrac{1}{n^3}\right)=\frac{\cosh \left(\frac{\sqrt{3} \pi }{2}\right) \Gamma \left(m-\frac{i \sqrt{3}}{2}+\frac{3}{2}\right) \Gamma \left(m+\frac{i \sqrt{3}}{2}+\frac{3}{2}\right)}{3 \pi m^3 \Gamma (m)^2}$$ and $$\prod_{n=2}^{m}\left(1+\dfrac{1}{n^3}\right)=\frac{(m+1) \cosh \left(\frac{\sqrt{3} \pi }{2}\right) \Gamma \left(m-\frac{i \sqrt{3}}{2}+\frac{1}{2}\right) \Gamma \left(m+\frac{i \sqrt{3}}{2}+\frac{1}{2}\right)}{2 \pi \Gamma (m+1)^2}$$ So, the product, from $n=2$ to $n=m$, write $$\frac{(m+1) \cosh ^2\left(\frac{\sqrt{3} \pi }{2}\right) \Gamma \left(m-\frac{i \sqrt{3}}{2}+\frac{1}{2}\right) \Gamma \left(m-\frac{i \sqrt{3}}{2}+\frac{3}{2}\right) \Gamma \left(m+\frac{i \sqrt{3}}{2}+\frac{1}{2}\right) \Gamma \left(m+\frac{i \sqrt{3}}{2}+\frac{3}{2}\right)}{6 \pi ^2 m^5 \Gamma (m)^4}$$ If $m$ goes to infinity, the limit is then $$\frac{1+\cosh \left(\sqrt{3} \pi \right)}{12 \pi ^2} =\dfrac{\cosh^2\left(\pi\dfrac{\sqrt3}2\right)}{6\pi^2}$$ as shown by Lucian.

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The connection between the Gammas of the odd exponents and the sines of the even ones is the result of Euler's reflection formula. – Lucian Feb 13 '14 at 6:39
@Lucian. Thank you for that since I learn a lot from you. I must say that I really like your answer. I thought that extending what was done on yesterday was also an interesting execrcise. Cheers. – Claude Leibovici Feb 13 '14 at 7:00

$$P_6(x)=\prod_{n=1}^\infty\bigg(1-\frac{x^6}{\pi^6n^6}\bigg)=\frac{\sin x}{x^3}\cdot\frac{\cosh\big(x\sqrt3\big)-\cos x}2\quad$$

See Basel problem. For $x\to\pi$, use l'Hopital after first dividing through $1-\dfrac{x^6}{\pi^6}$ , in order to finally arrive at $\dfrac{\cosh^2\left(\pi\dfrac{\sqrt3}2\right)}{6\pi^2}$ . In general, $P_{2k}(x)=s_k\displaystyle\prod_{j=0}^{k-1}\frac{\sin\Big((-1)^{j/k}x\Big)}x$ , where $s_k$ forms a cycle of length four, $(+,+,-,-)$, starting with $k=1$. And, of course, when $x\to\pi$, use l'Hopital's rule after first dividing through $1-\bigg(\dfrac x\pi\bigg)^{2k}$, just as before. Hope this helps.

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