Mean value of arithmetic function Suppose we define a mean value of arithmetic function $G(f)$ as $$ G(f)=\lim_{x \rightarrow \infty} \frac{1}{x \log{x}} \sum_{n \leq x} f(n) \log{n},$$ and suppose now for an arithmetic function $f$, $G(f)$ exist and is equal to $A$, how to use this result to show that the ordinary mean value of arithmetic function $M(f)=\lim_{x \rightarrow \infty} \frac{1}{x} \sum_{n \leq x} f(n)$ also exists?
 A: Via Abel's summation formula:
$$\sum_{n\le x} (f(n)\log n)\frac{1}{\log n}=\left(\sum_{n\le x}f(n)\log n\right)\frac{1}{\log x}+\int_2^x \left(\sum_{m\le u} f(m)\log m\right)\frac{du}{u\log^2 u}.$$
Divide by $x$ and subtract, obtain:
$$M_x(f)-G_x(f)=\frac{1}{x}\int_2^xG_u(f)\frac{du}{\log u}=\frac{\mathrm{Li}(x)}{x}\left(A+O(1)\right)\to0.$$
A: You'll want to use "partial summation", also called "summation by parts". Define $G(f;x) = \sum_{n\le x} f(n)\log n$ and $M(f;x) = \sum_{n\le x} f(n)$. Then you can write $M(f;x)$ as a Riemann-Stieltjes integral
$$
M(f;x) = \int_1^x \frac1{\log t} \, dG(f;t).
$$
(Technically the lower endpoint should be $1-\epsilon$.) Then integrating by parts gives
$$
M(f;x) = \frac{G(f;x)}{\log x} + \int_1^x \frac{G(f;t)}{t(\log t)^2} \,dt. \tag1
$$
(Even if you don't know Riemann-Stieltjes integrals, you can still verify this last identity by hand - just split the integral up into intervals of length 1, on which $G$ is constant.)
When you divide both sides of equation (1) by $x$ and take the limit as $x\to\infty$, all that remains to show is that the term with the integral tends to $0$. (Note that $G(f;t)=0$ for $t<2$, so there's no problem with the integral at the lower endpoint.)
