Evaluate these infinite products $\prod_{n\geq 2}(1-\frac{1}{n^3})$ and $\prod_{n\geq 1}(1+\frac{1}{n^3})$ What is


*

*$\prod\limits_{n\geq 2}(1-\frac{1}{n^3})=?$

*$\prod\limits_{n\geq 1}(1+\frac{1}{n^3})=?$


I am sure about their convergence. But don't know about exact values. Know some bounds as well. For example first one is in interval (2/3,1) and second one is in (2,3).
 A: *

*Ramanujan has already answered these two questions $100$ years ago!


*The same questions have already been answered here, by Mhenni Benghorbal, and here, by Claude Leibovici.


*Generally speaking, $\quad\displaystyle\prod_{n=1}^\infty\bigg(1\pm\frac{x^{2k+1}}{n^{2k+1}}\bigg)=\pm\frac1{x^{2k+1}\cdot\displaystyle\prod_{j=1}^{2k+1}\Gamma\bigg((\pm1)^{^{\frac j{2k+1}}}x\bigg)}$ . When $x=1$ and the sign is negative, in order to find the value of your product, you must obviously divide through $1-\bigg(\dfrac xn\bigg)^{2k+1}$ But since both the numerator and denominator are $0$, we must use l'Hopital. For $k=1\iff2k+1=3$, this will lead you to the above-linked results, which have also been confirmed by Gamma Tester in this post's comment section.


*As far as the relationship to the $\zeta(3)$ function is concerned, this is clear from Euler's approach to the Basel problem, which connects $\zeta(2)$ with the infinite product for $\dfrac{\sin x}x$ , as well as from the fact that, in general, the convergence of $\prod(1+a_k)$ is the same as that of $\sum a_k$. Hope this helps!
A: Just one quick observation. 
Let, $P_1$ denotes the first product and $P_2$ the second. Then,
since $P_1$ converges and is positive, $$\log{P_1}=\sum_{n\ge 2}\log\left(1-\frac 1{n^3}\right)=-\sum_{n\ge 2}\sum_{k\ge 1}\frac{1}{kn^{3k}}=-\sum_{k\ge 1}\frac{\zeta(3k)-1}{k}$$
Similarly, $$\log P_2=\sum_{k\ge 1}\frac{(-1)^{k-1}\zeta(3k)}{k}$$
