7
$\begingroup$

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?

$\endgroup$
0
5
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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.)

$\endgroup$
3
  • $\begingroup$ (+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 $A$. $\endgroup$
    – anon
    Oct 1 '11 at 9:40
  • $\begingroup$ True ... the integrand is in fact $(A+o(1))/\log t$, but the integral still turns out to be $o(1)$. $\endgroup$ Oct 2 '11 at 3:29
  • $\begingroup$ Oh yes, you're correct. $\endgroup$
    – anon
    Oct 2 '11 at 3:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.