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First I rewrite $\prod_{n=1}^{\infty}\left(1+\frac{1}{n^{3}}\right)$ as $\prod_{n=1}^{\infty}\left(\frac{1+n^{3}}{n^{3}}\right)$, then by factor out polynomial I get $\prod_{n=1}^{\infty}\left(\frac{(1+n)(n^{2}-n+1)}{n^{3}}\right)$ which is a problem because I can't factor any further which makes me stuck on this step I would hope for any help.

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  • $\begingroup$ Just an idea: take the logarithm and note that $$\log{\left (1+\dfrac1{n^3} \right )} = \int_0^1 \dfrac{dx}{n+x}-\int_0^1 \dfrac{(n-1)dx}{n^2-(n-1)x}.$$ By interchanging sum and integrals, we would get something involving digamma function. Hopefully this answer can "easily" express in-terms of hyperbolic cosine. $\endgroup$
    – Bumblebee
    Jan 27 '21 at 3:13
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If you're familiar with the gamma function, then this (or this... etc) is the way to go.

Otherwise, write $$1+\frac{1}{n^3}=\left(1+\frac1n\right)\left(1-\frac{1}{2n}\right)^2\left(1+\frac{3}{(2n-1)^2}\right)$$ and see that $$\prod_{n=1}^N\left(1+\frac1n\right)\left(1-\frac{1}{2n}\right)^2=(N+1)\left(\frac{(2N-1)!!}{(2N)!!}\right)^2=(N+1)\left(\frac{(2N)!}{2^{2N}N!^2}\right)^2$$ tends to $1/\pi$ as $N\to\infty$ (by Wallis product, or Stirling's formula), and $$\prod_{n=1}^\infty\left(1+\frac{3}{(2n-1)^2}\right)=\cosh\frac{\pi\sqrt3}{2}$$ is a special case of the infinite product for the [hyperbolic] cosine.

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  • $\begingroup$ I love this approach and I choose it as best one because it avoids the factorization involving complex number, I would like to thank others who have answered as well. $\endgroup$ Jan 27 '21 at 2:57
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[Note] : I shall use the following identites :

  • Identity $(1)$ : $\quad\displaystyle\left|\Gamma\left(\frac{1}{2}+\beta i\right)\right|=\frac{\pi}{\cosh(\pi\beta)}$
  • Identity $(2)$ :$\quad\displaystyle\lim_{n\to\infty}\frac{\Gamma(n+\alpha)}{\Gamma(n+\beta)}\approx n^{\alpha-\beta}$
  • Identity $(3)$ : $\quad\displaystyle\prod_{k=1}^{n}(x+k)=\frac{\Gamma(n+x+1)}{\Gamma(x+1)}$

We proceed as follow : \begin{align*} \prod_{n=1}^{\infty}\left(1+\frac{1}{n^{3}}\right)&=\prod_{n=1}^{\infty}\left(\frac{n^{3}+1}{n^{3}}\right)\\ \\ &=\prod_{n=1}^{\infty}\left[\frac{(1+n)(n^{2}-n+1)}{n^{3}}\right] \\ \\ &=\lim_{\gamma\to\infty}\prod_{n=1}^{\gamma}\left[\frac{(n+1)\left(n+\frac{-1+i\sqrt{3}}{2}\right)\left(n+\frac{-1-i\sqrt{3}}{2}\right)}{n^{3}}\right] \\ \\ &=\lim_{\gamma\to\infty}\frac{\Gamma(\gamma+2)\cdot\displaystyle\frac{\Gamma\left(\gamma+\displaystyle\frac{1+i\sqrt{3}}{2}\right)}{\Gamma\left(\frac{1+i\sqrt{3}}{2}\right)}\displaystyle\cdot\frac{\Gamma\left(\gamma+\frac{1-i\sqrt{3}}{2}\right)}{\Gamma\left(\frac{1-i\sqrt{3}}{2}\right)}}{\Gamma(\gamma+1)\Gamma(\gamma+1)\Gamma(\gamma+1)} \\ \\ &=\frac{1}{\displaystyle\underbrace{\Gamma\left(\frac{1+i\sqrt{3}}{2}\right)\Gamma\left(\frac{1-i\sqrt{3}}{2}\right)}_{\text{Reflection}}} \\ \\ &=\frac{1}{\displaystyle\frac{\pi}{\cosh\left(\frac{\pi\sqrt{3}}{2}\right)}} \\ \\ &=\frac{\cosh\left(\frac{\pi\sqrt{3}}{2}\right)}{\pi} \end{align*}

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Use $$\frac{e^{-\gamma z}}{\Gamma(1+z)}=\prod_{n=1}^{\infty} (1+\frac{z}{n})e^{-z/n}~~~~(1)$$ Next, $$\left(1+\frac{1}{n^3}\right)=\left(1+\frac{1}{n}\right) \left(1+\frac{w}{n}\right) \left(1+\frac{w^2}{n}\right),$$ where $w$ is cube root of unity $w^3=1, w^2+w+1=0, w^2=w^*$ $$P=\prod_{1}^{\infty} \left(1+\frac{1}{n^3}\right)$$ From (1), we get $$ e^{-(1+w+w^2)\gamma}[\Gamma(2)\Gamma(1+w)\Gamma(1+w^2)]^{-1}= [\Gamma(-w) \Gamma(-w^2)]^{-1}=\prod_{n=1}^{\infty} \left(1+\frac{1}{n^3}\right) e^{(1+w+w^2)/n}$$ So $$P=\frac{1}{|\Gamma(-w)|^2}=$$ $$P=\frac{1}{\Gamma[(1-i\sqrt{3})/2]~ \Gamma[(1+i\sqrt{3})/2]}$$ Use $\Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin \pi z}$ to evaluate $\Gamma(1/2-it) \Gamma(1/2+it)=\frac{\pi}{\cosh \pi t}$. Then the required $$P=\Gamma(1/2-it) \Gamma(1/2+it), t=\pi\sqrt{3}/2$$ Hence $$P=\frac{1}{\pi} \cosh(\sqrt{3}\pi/2).$$.

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  • $\begingroup$ Correction: the product in $(1)$ equals $e^{-\gamma z}/\Gamma(1+z)$, not just $1/\Gamma(z)$. $\endgroup$
    – metamorphy
    Jan 27 '21 at 7:36
  • $\begingroup$ @metamorphy Thanks, I have incorporated this correction. $\endgroup$
    – Z Ahmed
    Jan 27 '21 at 8:02

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