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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 $\alpha$?

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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

3 Answers 3

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 $\alpha>0$.

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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
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@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 $.

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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 $\alpha>0$.

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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

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