The Prime number theorem is equivalent to the argument$$\sum_{n\ge 1}\dfrac{\mu(n)}{n}=0$$ Is it possible to imply that $\displaystyle \sum_{n\ge 1}\dfrac{\mu(n)}{n}=\prod_{p \text { prime }}\left(1-\dfrac{1}{p}\right)$? If it is the case, then by Merten's theorem $$\prod_{p\text{ prime }}\left(1-\dfrac{1}{p}\right)=\lim_{x\rightarrow\infty}\prod_{p\le x}\left(1-\dfrac{1}{p}\right)=\lim_{x\rightarrow\infty}\dfrac{e^{-\gamma}}{\log x}=0$$
and therefore, the PNT. This recall me that it might not be the case. Another analogous one is that can we write $\displaystyle \sum_{n\ge 1}\dfrac{\mu(n)}{\sqrt n}=\prod_{p}\left(1-\dfrac{1}{\sqrt p}\right)$? And, as far as I concern $\displaystyle \sum_{n\ge 1}\dfrac{\mu(n)}{n^\alpha}$ converges if $\alpha >1$, even when $\alpha =1$ as it is equivalent to PNT. I wonder if $\displaystyle \sum_{n\ge 1}\dfrac{\mu(n)}{\sqrt n}$ converges as well, or is there any hint to tackle with? I would really appreciate for any help, thank you.