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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?

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Punctution should go inside double dollar signs (ideally separated by "\;"), otherwise it appears on the next line. –  joriki Oct 1 '11 at 8:46
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2 Answers

up vote 4 down vote accepted

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: $$M_x(f)-G_x(f)=\frac{1}{x}\int_2^xG_u(f)\frac{du}{\log u}=\frac{\mathrm{Li}(x)}{x}\left(A+O(1)\right)\to0.$$

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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.)

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(+1) I just realized my answer is essentially the same thing, I didn't see it at first because I'm not used to RS integration. The integrand in $\mathrm{(1)}$ is $(A+o(1))/t$ so its integral divided by $x$ is $\frac{\log x}{x}(A+O(1))$ which is $o(1)$. Thus the arithmetic mean exists and equals $A$. –  anon Oct 1 '11 at 9:40
    
True ... the integrand is in fact $(A+o(1))/\log t$, but the integral still turns out to be $o(1)$. –  Greg Martin Oct 2 '11 at 3:29
    
Oh yes, you're correct. –  anon Oct 2 '11 at 3:43
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