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Ramanujan's notebooks contain the result $$\prod_{k=1}^{\infty} \Big( 1 + \frac{1}{k^3}\Big) = \frac{1}{\pi} \mathrm{cosh}\Big( \frac{\pi \sqrt{3}}{2}\Big).$$ It doesn't seem like this is proved there but I guess it is well-known. How could you show this?

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marked as duplicate by Michael Albanese, This is much healthier., Kirill, Hans Lundmark, Claude Leibovici Jul 5 '14 at 7:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Sorry if this has been asked before. I didn't find it. – user161722 Jul 5 '14 at 3:07
See here and here. – Lucian Jul 5 '14 at 3:20
up vote 4 down vote accepted

Since the Weierstrass product for the Gamma function gives: $$\frac{1}{\Gamma(z)} = z\, e^{\gamma z}\cdot \prod_{k=1}^{+\infty}\left(1+\frac{z}{k}\right)e^{-\frac{z}{k}}$$ given that $\omega$ is a primitive third root of unity, we have: $$\prod_{k=1}^{+\infty}\left(1+\frac{z^3}{k^3}\right)=\frac{1}{z^3 \Gamma(z)\Gamma(\omega z)\Gamma(\omega^2 z)}=\frac{1}{\Gamma(z+1)\Gamma(\omega z+1)\Gamma(\omega^2z+1)},$$ hence: $$\prod_{k=1}^{+\infty}\left(1+\frac{1}{k^3}\right)=\frac{1}{\Gamma(\omega+1)\Gamma(\omega^2+1)},$$ but since $(\omega+1)+(\omega^2+1)=1$ and $$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)},$$ it follows that: $$\prod_{k=1}^{+\infty}\left(1+\frac{1}{k^3}\right)=\frac{1}{\pi}\sin(\pi(\omega+1))=\frac{1}{\pi}\sin\left(\frac{\pi}{2}+i\frac{\pi\sqrt{3}}{2}\right)=\frac{1}{\pi}\cosh\frac{\pi\sqrt{3}}{2}.$$

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