About the limit of $\prod\limits_{k=2}^n (1-\frac{1}{k^3})^k (1+\frac{1}{k^3})^{1-k}$ Question: 
Does a formula exist for 
$\enspace\displaystyle \prod\limits_{k=2}^n (1-\frac{1}{k^3})^k (1+\frac{1}{k^3})^{1-k}\enspace$ so that it can be seen 
that the limit is $\,\displaystyle\frac{1}{3}\,$ for $\,n\to\infty\,$ ? 
Note: 
The input $\enspace\text{prod (1-1/k^3)^k/(1+1/k^3)^(k-1) from k=2 to n}$ 
with Wolfram Alpha gives a complicated term, with which I cannot see any limit.
Also to take the logarithm doesn't look very helpful.

EDIT: I thank all for the kind and competent help.
 A: Well, $\prod_{k\geq 2}\left(1+\frac{1}{k^3}\right)=\frac{1}{2\pi}\cosh\left(\frac{\pi\sqrt{3}}{2}\right)$, hence we may focus on
$$ \prod_{k\geq 2}\left[\frac{1-\frac{1}{k^3}}{1+\frac{1}{k^3}}\right]^k=\exp\sum_{k\geq 2}-2k\,\text{arctanh}\frac{1}{k^3}=\exp\sum_{k\geq 2}\sum_{n\geq 0}\frac{-2}{(2n+1)k^{6n+2}} $$
which can be written as
$$ \exp\sum_{n\geq 0}\frac{2\left[1-\zeta(6n
+2)\right]}{(2n+1)}=\exp\sum_{n\geq 0}\frac{-2}{2n+1}\int_{0}^{+\infty}\frac{x^{6n+1}}{(6n+1)!e^x(e^x-1)}\,dx$$
or as
$$ \exp\int_{0}^{+\infty}\frac{-g(x)}{e^x(e^x-1)}\,dx $$
where $g(x)$ is a hypergeometric function and $\frac{g(x)}{e^x(e^x-1)}$ behaves like $\left(2-\frac{x^2}{12}\right) e^{-3x/2}$ in a right neighbourhood of the origin. This leads to the approximation $\approx 0.336233$ for the original product. As shown by Batominovski, the original product is exactly $\frac{1}{3}$ by telescoping. This leads to the identity
$$ \sum_{n\geq 0}\frac{\zeta(6n+2)-1}{2n+1}=\frac{1}{2}\log\left(\frac{3}{2\pi}\cosh\frac{\pi\sqrt{3}}{2}\right)\tag{I}$$
and to the approximate identity
$$ \cosh\left(\frac{\pi\sqrt{3}}{2}\right)\approx \frac{2\pi}{3}e^{\frac{104}{81}}.\tag{II}$$
A: You can prove via induction or telescopic product that
$$\prod_{k=2}^n\,\left(1-\frac1{k^3}\right)^k\,\left(1+\frac1{k^3}\right)^{1-k}=\frac{\left(n^2+n+1\right)^n}{3\,n^{n+1}\,(n+1)^{n-1}}$$
for all $n=2,3,4,\ldots$.
A: You can have a good approximation taking logarithms since
$$a_k=k \log \left(1-\frac{1}{k^3}\right)+(1-k) \log \left(1+\frac{1}{k^3}\right)$$ Using Taylor expansions
$$a_k=-\frac{2}{k^2}+\frac{1}{k^3}-\frac{1}{2 k^6}-\frac{2}{3
   k^8}+\frac{1}{3 k^9}+O\left(\frac{1}{k^{12}}\right)$$
$$S_n=\sum_{k=2}^n a_k=\frac{1}{6} \left(-12 H_n^{(2)}+6 H_n^{(3)}-3 H_n^{(6)}-4 H_n^{(8)}+2
   H_n^{(9)}+11\right)$$ where appear generalized harmonic numbers.
Using asymptotics, this would give
$$S_n=\left(\zeta (3)+\frac{\zeta (9)}{3}+\frac{11}{6}-\frac{\pi ^2 \left(9450+15 \pi
   ^4+2 \pi ^6\right)}{28350}\right)+\frac{2}{n}+O\left(\frac{1}{n^2}\right)$$ Taking the exponential of the first term will give $0.333360$
