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. 



asymptotic for li(x)-Ri(x)
Yes, the stated asymptotics (and much more) is true. The idea is to truncate $\operatorname{Ri}(x)$ appropriately. 


Let us use the series representation (see [here](https://en.wikipedia.org/wiki/Logarithmic_integral_function#Series_representation))
$$\operatorname{li}(t)=\gamma+\log\log t+\sum\_{k=1}^\infty\frac{(\log t)^k}{k\cdot k!},\qquad t>1.$$
This implies
$$\operatorname{li}(t)=\gamma+\log\log t+O(\log t),\qquad 1<t<e,$$ 
hence also
$$\operatorname{li}(x^{1/n})=\gamma+\log\log x-\log n+O\left(\frac{\log x}{n}\right),\qquad n>\log x.$$
As a result, for $x>3$ we get
\begin{align\*}\sum\_{n>\log x} \frac{\mu(n)}{n} \operatorname{li}(x^{1/n})&=O(1)+\sum\_{n>\log x}\frac{\mu(n)}{n}\left(\gamma+\log\log x-\log n\right)\\[6pt]
&=O\left((\log\log x)^2\right)+\sum\_{n=1}^\infty\frac{\mu(n)}{n}\left(\gamma+\log\log x-\log n\right)\\[6pt]
&=O\left((\log\log x)^2\right).\end{align\*}
Here we used that
$$\sum\_{n=1}^\infty\frac{\mu(n)}{n}=0\qquad\text{and}\qquad\sum\_{n=1}^\infty\frac{\mu(n)\log n}{n}=-1$$
by the prime number theorem. To summarize so far,
$$\operatorname{Ri}(x)=\sum\_{n\leq\log x} \frac{\mu(n)}{n} \operatorname{li}(x^{1/n})+O\left((\log\log x)^2\right),\qquad x>3.$$
The error term on the right hand side can probably be improved, but this is not important. In the sum, we detach the first three terms (corresponding to $n=1,2,3$) and estimate the rest trivially. We get
$$\operatorname{Ri}(x)=\operatorname{li}(x)-\frac{1}{2}\operatorname{li}(x^{1/2})+O\left(\frac{x^{1/3}}{\log x}\right),\qquad x>3.$$


