Is it true that $$\operatorname{li}(x)-\operatorname{Ri}(x) \sim \frac{1}{2}\operatorname{li}(x^{1/2}) \ (x \to \infty),$$ where $$\operatorname{Ri}(x) = \sum_{n = 1}^\infty \frac{\mu(n)}{n} \operatorname{li}(x^{1/n}) = 1 + \sum_{k = 1}^\infty \frac{(\log x)^k}{k \cdot k!\ \zeta(k+1)}$$ for all $x > 0$? If so, how can one prove the given asymptotic?

Note that \begin{align}\label{lirieq} \lim_{x \to \infty} \frac{\operatorname{li}(x) - \operatorname{Ri}(x)}{\frac{1}{2}\operatorname{li}(x^{1/2})} = 1- 2\lim_{x \to \infty} \sum_{n = 3}^\infty \frac{\mu(n)}{n}\frac{\operatorname{li}(x^{1/n})}{\operatorname{li}(x^{1/2})} = 1-2\sum_{n = 3}^\infty \lim_{x \to \infty} \frac{\mu(n)}{n}\frac{\operatorname{li}(x^{1/n})}{\operatorname{li}(x^{1/2})} = 1, \end{align} provided that the given limit can be interchanged with the given sum. However, I am unable to justify interchanging the limit with the sum.