I define for some set of real numbers $x\in S$ (see that it is my **Question 1.**) the domain of the function $$f(x)=\prod_{k=2}^\infty \left( 1+\frac{\mu(k)}{k^3} x\right) ,$$
where $\mu(k)$ is the **Möbius function.**

Question 1.I am interesting in this function since I believe that it is possible justify $f'(0)=\frac{1}{\zeta(3)}-1$. If it is well known in the literature or you know how compute its domain $S$, it is for which $x$ is defined, can you tell me?

My calculations were that $0<\frac{7}{8}\leq 1+\frac{\mu(k)}{k^3}<1$ for integers $k\geq 2$, and I am stuck with what theorem relating infinite products and series can I use to answer previous question.

Subsequents calculations provide to me, informally $$\log f(1)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\sum_{k=2}^\infty\frac{(\mu(k))^n}{k^{3n}}.$$

When I've splitted the series for odd and even $n's$, I can write $\log f(1)$ as the sum two terms, the first $$\sum_{\text{n odd}}\frac{(-1)^{n+1}}{n}\sum_{k=2}^\infty\frac{(\mu(k))^n}{k^{3n}}=\sum_{j=1}^\infty\frac{\zeta(3(2j-1)-1)}{2j-1},$$ and the second as $$\sum_{\text{n even}}\frac{(-1)^{n+1}}{n}\sum_{k=2}^\infty\frac{(\mu(k))^n}{k^{3n}}=-\sum_{j=1}^\infty\frac{1}{2j}\sum_{k=2}^\infty\frac{ \left| \mu(k) \right| }{k^{6j}}.$$

Question 2.Were justified my calculations for $\log f(1)$? Only is requiered a yes or where was my mistake. Also if my calculations were right if you know how write the series inside $$-\sum_{j=1}^\infty\frac{1}{2j}\sum_{k=2}^\infty\frac{ \left| \mu(k) \right| }{k^{6j}}$$ with zeta values, as I've computed for the case $\text{ n odd}$, then these nice computations will be here as reference for all us.Thanks in advance.