Exponentiating to get the gaussian $e^{-x^2}$, for $0<x<1$ from the infinite product that satisfies the Möbius function, see the identity (16) from the article Möbius Function from MathWolrd, then I've asked myself
Question 1. Can be proven that the sequence of partial products $$ \left\{ \prod_{k=1}^n \left( 1-x^k \right)^{\frac{\mu(k)}{k}x} \right\}_{n\geq 1} $$ is uniformly convergent to our gaussian $e^{-x^2}$ on $(0,1)$? What are your hints? Many thanks.
I say a proof without using explicitly the identity that I've evoked as main argument in the proof. I believe that our sequence of partial products is uniformly convergent on compact subsets of $(0,1)$, what I would like to learn is how to formalize this claim, I remember Cauchy condition.
I did a second set of calculations inspired in the fact that $\int e^{-x^2}dx=\frac{\sqrt{\pi}}{2}\operatorname{erf}(x)+\operatorname{constant}$. Combining the Prime Number Theorem with the MacLaurin series expansion of the Error function $\operatorname{erf}(z)$ defined from this MathWorld article for Erf (the examples seems also feasibles for $\operatorname{erfi}(z)$ ). Combining also with a related Dirichlet series. Then walking this way I can deduce identities whose series in my RHS have terms involving particular values of the Riemann Zeta function, for example for $$\frac{\sqrt{\pi}}{2}\sum_{k=1}^\infty\lambda(k)\operatorname{erf}\left(\frac{1}{k}\right)$$ where $\lambda(n)$ is the Liouville function. I would like to know some rigurous proof or reasoning to convice us that previous series is convergent. I believe that the series in our LHS for each of our examples, doesn't converge absolutely.
Question 2. Please, what is your explanation or hints to show that $$\frac{\sqrt{\pi}}{2}\sum_{k=1}^\infty\lambda(k)\operatorname{erf}\left(\frac{1}{k}\right)$$ does converge? Thanks and good wishes.
My reasoning to get a different example. Since the example seems also feasible with the Möbius function instead Liouville function, and same error function, my justification for this different example was using the ratio test in RHS $$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n} \right|=\lim_{n\to\infty}\frac{2n+1}{(n+1)(2n+1)}\lim_{n\to\infty}\zeta(2n+1)\lim_{n\to\infty}\frac{1}{\zeta(2n+1)}=0\cdot1\cdot1=0<1,$$ and thus our series converges, but as I am saying I would like to know your calculations and reasonings for previous different example in Question 2, to do a comparison.
Let us consider $$f\left(x\right)=-\sum_{k\geq1}\frac{\mu\left(k\right)}{k}\log\left(1-x^{k}\right). $$ Now for the Möbius inversion for the power series we get $$\sum_{k\geq1}\frac{f\left(x^{k}\right)}{k}=-\log\left(1-x\right) $$ hence $f\left(x\right)=x $ and so $$\prod_{k\geq1}\left(1-x^{k}\right)^{x\mu\left(k\right)/k}=\exp\left(x\sum_{k\geq1}\frac{\mu\left(k\right)}{k}\log\left(1-x^{k}\right)\right)=\exp\left(-x^{2}\right). $$ For the second we may use $$\textrm{erf}\left(x\right)\sim\frac{2x}{\sqrt{\pi}},\, x\rightarrow0 $$ hence $$\lambda\left(k\right)\textrm{erf}\left(\frac{1}{k}\right)\sim\frac{2}{\sqrt{\pi}}\frac{\lambda\left(k\right)}{k} $$ and recalling that $$\sum_{k\geq1}\frac{\lambda\left(k\right)}{k^{s}}=\frac{\zeta\left(2s\right)}{\zeta\left(s\right)}$$ we can conclude.