1
$\begingroup$

Let $\chi$ be a nontrivial Dirichlet character modulo $m$. I am reading the argument in Davenport where he proves that $\sum_p \frac{\chi(p)}{p}$ converges.

I can establish after some work that $\sum_{p \leq x} \frac{\chi(p) \log p}{p} = O(1)$. I'm looking for someone to spell out if I can get from this to proving that $\sum_p \frac{\chi(p)}{p}$ converges.

$\endgroup$
1
$\begingroup$

Sure. Let $\Lambda(n) = \log p$ if $n=p^k$, $\Lambda(n) = 0$ otherwise, and $$\rho(N,\chi) = \sum_{n \le N} \frac{\chi(n)\Lambda(n)}{n} $$

From $\rho(N,\chi)= \mathcal{O}(1)$ for $\chi$ non-principal (a consequence of $L(1+it,\chi) \ne 0$)

and since $$\frac{1}{\log n}-\frac{1}{\log (n+1)} = \int_n^{n+1} \frac{dx}{x \log^2 x}= \mathcal{O}(\frac{1}{n \log^2 n})$$

using partial summation we obtain $$\sum_{n=2}^N \frac{\chi(n) \Lambda(n)}{n\log n} = \frac{\rho(N,\chi)}{\log N} + \sum_{n=2}^{N-1} \rho(n,\chi) (\frac{1}{\log n}-\frac{1}{\log (n+1)}) \\= \mathcal{O}(\frac{1}{\log N}) + \sum_{n=2}^{N-1} \mathcal{O}(\frac{1}{n \log^2 n}) = C(\chi)+\mathcal{O}(\frac{1}{ \log N})$$ Where $$C(\chi) = \sum_{n=2}^\infty \rho(n,\chi) (\frac{1}{\log n}-\frac{1}{\log (n+1)})\qquad (= \log L(1,\chi))$$

$\endgroup$
  • $\begingroup$ Can you say what is happening in the equality $O(1/log N) + \sum_{n=2}^N O(\frac{1}{n\log^2 n})$? If we skip this equality and go to the next, I see that the series defining $C(\chi)$ converges, just because $\rho$ is bounded and using this bound we have a telescoping series for the log's and hence get an upper bound of $1/\log 2$. From this I get that the series $\sum_{n=2}^\infty \frac{\chi(n) \Lambda(n)}{n \log n}$ converges. How am I doing? $\endgroup$ – Jordan Nov 19 '17 at 1:33
  • 1
    $\begingroup$ @Jordan Sure you can use that the sum is telescoping, but I wanted to use $\frac{1}{\log n}-\frac{1}{\log (n+1)} = \int_n^{n+1} \frac{dx}{x \log^2 x}= \mathcal{O}(\frac{1}{n \log^2 n})$ showing how Abel summation formula and summation by parts are the same thing, and which works even when $\rho$ is unbouned $\endgroup$ – reuns Nov 19 '17 at 1:46
  • $\begingroup$ Also the $\mathcal{O}$ constants depend on the one for $\rho(N,\chi)= \mathcal{O}(1)$, thus I should have written $\mathcal{O}_\chi$ everywhere $\endgroup$ – reuns Nov 19 '17 at 1:48

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.