MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have been reading a paper on the Goldbach conjecture found at At one point, the author (Paul Truman), states: Let $z=N^{1/8}$, then $$\sum_{w\leq z}\frac{d(w)}{w}\gg(\log(z))^2\gg(\log N)^2$$ where $d(w)$ counts all the positive divisors of $w$. I am assuming that there's a mistake in the second part of the inequality $(\log(z))^2\gg(\log N)^2$, but this is not the first time I've encountered such a claim: at the author claims (in his proof of the upper bound on the twin prime counting function) that $$\sum_{\substack{d\leq N^{1/3}\\ d \\ odd}}\frac{f(d)}{d} \geq (\log N)^2$$ where $f(2)=1, f(p)=2$ for all odd primes $p$. This inequality I have seen proved (though I can't recall where) by saying that $f(w)/w\geq d(w)/w$. A lot of people are saying that $$\sum_{w\leq z}\frac{d(w)}{w}\gg(\log(z))^2$$ but others are saying that $$\sum_{w\leq z}\frac{d(w)}{w}=\frac{\log^2(z)}{2}+O(\log x)$$ including how can both be true?

share|cite|improve this question
Do you know what the $\gg$ notation means? This may be where you're getting confused. In particular, if $x = N^{1/8}$, then it IS true that $\sum_{n \leq x}{\frac{d(n)}{n}} = \frac{1}{2} (\log x)^2 + O(\log x) \gg (\log x)^2 \gg (\log N)^2$. The notation $f(x) \gg g(x)$ usually means that for there exists $C > 0$ and $x_0 \in \mathbb{R}$ such that for all $x > x_0$, $f$ and $g$ are nonnegative and the inequality $f(x) \geq C g(x)$ holds. – Peter Humphries May 24 '11 at 3:28
It seems, at the very least, that you are confused about the Vinogradov notation $\gg$. In the context of your question, I think that one writes $A(x) \gg B(x)$ if there exists a constant $C > 0$ such that $A(x) \geq C B(x)$ for all $x$ under consideration (presumably, $x > 10$ or something like that). – Matt Young May 24 '11 at 3:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.