Compute $\lim\limits_{n\to\infty} \prod\limits_2^n \left(1-\frac1{k^3}\right)$ I've just worked out the limit $\lim\limits_{n\to\infty} \prod\limits_{2}^{n} \left(1-\frac{1}{k^2}\right)$ that is simply solved, and the result is $\frac{1}{2}$. After that, I thought of calculating $\lim\limits_{n\to\infty} \prod\limits_{2}^{n} \left(1-\frac{1}{k^3}\right)$, but I don't know how to do it. According to W|A, the result is pretty nice, but I don't see how W|A gets that. (See here.) Is there any easy way to get the answer?
 A: The last step of Andrew getting 
\begin{align}\lim_{n\to \infty}\prod _{k=2}^n \left(1-\frac{1}{k^3}\right)= \frac{\cosh \frac{\sqrt{3} \pi }{2} \Gamma \left(n-\frac{i \sqrt{3}}{2}+\frac{3}{2}\right) \Gamma \left(n+\frac{i \sqrt{3}}{2}+\frac{3}{2}\right)}{3 \pi n^3 \Gamma^2 (n)}\end{align} 
was a bit ambigous.
using another method, note that 
\begin{align*}\Gamma(z)=\frac{1}{z e^{\gamma z}}\prod_{k=1}^{\infty}\frac{k e^{\frac{z}{k}}}{z+k}
\end{align*}
holds for all complex number $z$ except negative integer, we obtain
\begin{align}g(z)=\prod_{k=1}^{\infty} (1+\frac{z}{k})e^{\frac{-z}{k}}=\frac{1}{z\Gamma(z)e^{\gamma z}}\end{align}
Thus
\begin{align}
g(\omega)g(\omega^2)=\prod_{k=1}^{\infty}\frac{k^2+k+1}{k^2}e^{-\frac{1}{k}}=\frac{1}{\Gamma(\omega)\Gamma(\omega^2) e^{\gamma}}=\frac{3}{e}\prod_{k=2}^{\infty}\frac{k^2+k+1}{k^2}e^{-\frac{1}{k}}
\end{align}
where $-\omega$ is the root of $x^3=1$
From
\begin{align}
\prod_{k=2}^{\infty}(1-\frac{1}{k})e^{\frac{1}{k}}=\lim_{n\to \infty}\frac{1}{n} e^{\frac{1}{2}+\cdots+\frac{1}{n}}=e^{\gamma -1}
\end{align}
Thus
\begin{align}
\prod_{k=2}^{\infty}\left(1-\frac{1}{k^3}\right)=\prod_{k=2}^{\infty}\frac{k^2+k+1}{k^2}e^{-\frac{1}{k}}\prod_{k=2}^{\infty}(1-\frac{1}{k})e^{\frac{1}{k}}=\frac{1}{3\Gamma(\omega)\Gamma(\omega^2)}
\end{align}
and hence the result
By the similar way we may get $\prod_{k=2}^{\infty}(1-\frac{1}{k^n})$
A: Since
$$
1-\frac1{k^3}=\frac{(k-1)(k+\frac12+\frac{\sqrt3}2i)(k+\frac12-\frac{\sqrt3}2i)}{k^3}
$$
and
$$
k+a=\frac{\Gamma(k+a+1)}{\Gamma(k+a)},
$$
every term in the product is a ratio of the Gamma functions. Also there is a formula
$$
\Gamma \left(\frac{1}{2}-i y\right) \Gamma \left(\frac{1}{2}+i y\right)=
\pi  \text{sech}\pi  y.
$$
In particular for the end terms of the product
$$\frac{1}{\Gamma \left(\frac{1}{2}+\frac{i \sqrt{3}}{2}\right) 
\Gamma \left(\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)}=\frac{\cosh \frac{\sqrt{3} \pi }{2}}{\pi }.
$$
Multiplying those ratios and canceling out the same terms leads to a formula for the partial product:
$$
\prod _{k=2}^n \left(1-\frac{1}{k^3}\right)=
\frac{\cosh \frac{\sqrt{3} \pi }{2} \Gamma \left(n-\frac{i
   \sqrt{3}}{2}+\frac{3}{2}\right) \Gamma \left(n+\frac{i
   \sqrt{3}}{2}+\frac{3}{2}\right)}{3 \pi  n^3 \Gamma^2 (n)}.
$$
Taking the limit $n\to\infty$ gives the desired result.
