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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.

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2 Answers 2

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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) $$\operatorname{li}(t)=\gamma+\log\log t+\sum_{k=1}^\infty\frac{(\log t)^k}{k\cdot k!},\qquad t>1.$$ 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 $$\operatorname{Ri}(x)=\operatorname{li}(x)-\frac{1}{2}\operatorname{li}(x^{1/2})+O\left(\frac{x^{1/3}}{\log x}\right),\qquad x>3.$$

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    $\begingroup$ Thanks! Your proof yields the stronger fact that the defining series expansion for $\operatorname{Ri}(x)$ is also an asymptotic expansion for $\operatorname{Ri}(x)$ at $\infty$. I suspected this to be true and am glad that your proof confirms it. Strangely I couldn't find a proof of this anywhere in the literature. $\endgroup$ Aug 19, 2018 at 21:07
  • $\begingroup$ Thanks for accepting my answer, I am glad I could help! $\endgroup$
    – GH from MO
    Aug 19, 2018 at 21:59
  • $\begingroup$ I'd like to provide this proof in a paper of mine. For now I've cited "GH from MO" and this link. If there is an alternative way you'd like me to cite this you can email me at Jesse.elliott@csuci.edu. $\endgroup$ Aug 21, 2018 at 23:39
  • $\begingroup$ @JesseElliott: I sent you an email. $\endgroup$
    – GH from MO
    Aug 22, 2018 at 12:42
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Since we have $${\rm Ri}(x)=1+\sum_{k\ge 1}\frac{(\log x)^k}{k\cdot k!\zeta(k+1)},$$ which is convergence absolutely for all $x>0$. Therefore we can use the L'Hospital's rule, and by $$\sum_{n\ge 1}\frac{\mu(n)}{n}=0$$ we have
\begin{align} \lim_{x\rightarrow +\infty}\frac{{\rm li}(x)-{\rm Ri}(x)}{\frac{1}{2}{\rm li}(x^{1/2})}&=\lim_{x\rightarrow +\infty}\frac{\frac{1}{\log x}-\frac{1}{x\log x}\sum_{k\ge 1}\frac{(\log x)^k}{k!\zeta(k+1)}}{\frac{1}{2}\frac{1}{\log(x^{1/2})}\cdot\frac{1}{2}x^{-1/2}}\\ &=\lim_{x\rightarrow +\infty}\left(2\sqrt{x}\left(1-\frac{1}{x}\sum_{k\ge 1}\frac{(\log x)^k}{k!}\sum_{n\ge 1}\frac{\mu(n)}{n^{k+1}}\right)\right)\\ &=\lim_{x\rightarrow +\infty}\left(2\sqrt{x}\left(1-\frac{1}{x}\sum_{n\ge 1}\frac{\mu(n)}{n}(x^{1/n}-1)\right)\right)\\ &=1-2\lim_{x\rightarrow +\infty}\left(\frac{1}{\sqrt{x}}\sum_{n\ge 3}\frac{\mu(n)}{n}(x^{1/n}-1)\right). \end{align} On the other hand, for $x$ sufficiently large, $$\sum_{3\le n\le x}\frac{\mu(n)}{n}(x^{1/n}-1)\ll x^{1/3}\log x$$ and $$\sum_{n>x}\frac{\mu(n)}{n}(x^{1/n}-1)\ll \sum_{n>x}\frac{1}{n}\left(\exp\left(\frac{\log x}{n}\right)-1\right)\ll \sum_{n>x}\frac{\log x}{n^2}\ll 1.$$ Hence we get $$\lim_{x\rightarrow +\infty}\left(\frac{1}{\sqrt{x}}\sum_{n\ge 3}\frac{\mu(n)}{n}(x^{1/n}-1)\right)=0.$$ Which completes the proof.

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