Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Where can I read about convergence of series constituted of prime number such as the following: $$\sum_p \frac{1}{p (\log{p})^\alpha}\;?$$ How does convergence depend on $\alpha$?

share|cite|improve this question
Punctuation should go inside double dollar signs (with a \; space); otherwise it gets placed on the next line. – joriki Sep 27 '11 at 8:29

Let $p_n$ be the $n$-th prime. By the Prime Number Theorem, $$ p_n\sim n\log n\;. $$ It follows that $$ \sum_n\frac{1}{p_n(\log p_n)^\alpha} $$ converges if and only if $\alpha>0$.

share|cite|improve this answer
So what did you use to conclude the result that you have (convegences if and only if $\alpha>0$)? – Rob Sep 27 '11 at 19:08
@Rob Comparison with the series $\sum\frac{1}{n(\log n)^p}\;,$ which converges if and only if $p>1$ (with $p=1+a$). – Julián Aguirre Sep 27 '11 at 19:54

I tried using Apostol's Introduction to Analytic Number Theory.

Theorem $ 4.12 $ on page $ 90 $ gives an asymptotic formula for $$ \sum_{p \leq x} \frac{1}{p}. $$ Using this and Theorem $ 4.2 $ on page $ 77 $ (Abel's Identity) with \begin{equation} a(n) := \left\{ \begin{array}{ll} \frac{1}{n} & \text{if $ n $ is prime;} \\ 0 & \text{otherwise} \end{array} \right. \end{equation} (so as to take the sum over all integers $ n $) and $ f(n) := \dfrac{1}{\log^{\alpha} n} $, I think you get that the partial sum is $ O(1) + O \left( \dfrac{1}{\log^{\alpha} x} \right) $, hence, convergence when $ \alpha > 0 $.

share|cite|improve this answer
The sum diverges for $\alpha = 0$. – JavaMan Oct 9 '11 at 3:20

Recall that $$S(x)=\sum_{p\leq x} \frac{1}{p}=\log \log x+B_1+E(x)$$ where $E(x)=O \left(e^{-c\sqrt{\log x}}\right).$ Then

$$\sum_{p\leq x} \frac{1}{p\log^\alpha p}=\int_2^x \frac{1}{\log^\alpha t}d(S(t))=\int_2^x \frac{1}{t\log^{\alpha+1} t}dt+\int_2^x \frac{1}{\log^\alpha t}d(E(t)).$$

By using partial summation you can prove that the second term will contribute very little. This means that the sum will be close to the main term $$\int_2^x \frac{1}{t\log^{\alpha+1}t}dt$$ and you can then prove that $$\sum_{p} \frac{1}{p\log^\alpha p}\ \ \text{converges} \iff \ \int_2^\infty \frac{1}{t\log^{\alpha+1}t}dt\ \text{converges}.$$ This happens for all $\alpha>0$.

share|cite|improve this answer
Eric, could you tell me where I can find a proof of your first relation? I know that this is Mertens type theorem, but in proofs that I know there is $E(x) = O(1/\log x)$. – xen Sep 27 '11 at 9:46

Your Answer


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

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