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.
Answer to Question 1
First, clearly for every $x$ there exists some $N\in \mathbb{N},$ such that $\frac{|x|}{k^3}<1$ for every $k>N.$
Since both $\sum_{k=2}^{\infty}\frac{\mu(k)}{k^3}x$ and $\sum_{k=2}^{\infty}\frac{|\mu(k)|^2 x^2}{k^6}$ converge, using the convergence criterion of infinite products, we can see that the infinite product definition of $f(x)$ converges for all $x\in\mathbb{R}.$
Thus, domain of $f(x)$ is the entire $\mathbb{R}.$
For detailed account of the convergent criterion, see criterion.
To calculate the derivative of a function $f$, we can take the derivative of its logarithm, $$ (\log f)^\prime=\frac{f^\prime}{f}. $$
Let $$ P_n=\prod_{k=2}^{n}\left (1+\frac{\mu(k)}{k^3}x\right). $$
We have \begin{align*} \log f(x)&=\log\lim\limits_{n\to\infty}P_n\\&=\lim\limits_{n\to\infty}\log P_n. \end{align*}
The first line is due to the definition of $P_n$ while the second line is due to the continuity of $\log x$.
Using Weierstrass M-test, it can be shown that both $\log P_n$ and $\frac{d}{dx}\log P_n$ converge uniformly, so we can take the derivative term by term, $$ (\log f(x))^\prime=\sum_{k=2}^{\infty}\frac{d}{dx}\log\left(1+\frac{\mu(k)}{k^3}x\right)=\sum_{k=2}^{\infty}\frac{\frac{\mu(k)}{k^3}}{1+\frac{\mu(k)}{k^3}x}. $$
In conclusion, we have \begin{align*} f^\prime(0)=f(0)(\log f)^\prime(0)=\sum_{k=2}^{\infty}\frac{\mu(k)}{k^3}=\frac{1}{\zeta(3)}-1. \end{align*}
For the Weierstrass M-test part, clearly for every $x$ there exists some $L\in \mathbb{N},$ such that $\frac{|x|}{k^3}<\frac{1}{2}$ for every $k>L.$
For $|x|<\frac{1}{2}$, we have
\begin{align*} |\log(1+x)|&\le |x|+|x|\bigl|\sum_{n\ge 2}\frac{(-1)^{n+1}x^{n-1}}{n}\bigr|\\ &\le |x|+|x||\sum_{n\ge 1}2^{-n}|\\ &\le 2|x|, \end{align*}
and \begin{align*} |\frac{1}{1+x}|<\frac{1}{1-|x|}<2. \end{align*}
For $k>L,$ we have
$$\biggl|\log\left (1+\frac{\mu(k)}{k^3}x\right)\biggr|<\frac{2|x|}{k^3},$$
and $$\biggl|\frac{d}{dx}\log\left (1+\frac{\mu(k)}{k^3}x\right)\biggr|<\frac{2}{k^3}.$$
Using these two estimates as the respective $M_k$ for $k>L$ and considering that adding $L$ finite terms doesn't change the convergence property of a series, we can see that both $\log P_n$ and $\frac{d}{dx}\log P_n$ converge uniformly for any finite interval.