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I wonder if there is some asymptotics for such sum: $ \sum_{p=2}^{n} \frac{1}{p}$, where the sum is taken over all primes of form $ 4k+3 $?

It's well-known that $ \sum_{p=2}^{n} \frac{1}{p}$, where the sum is taken over all primes is diverges and asymptotically is like $ \ln\ln n $.

But I don't even know how to prove that the first sum is diverges.

I am also interested in asymptotics of the number of primes of the form $ 4k+3 $ less than $n$.

Thank you very much!

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    I think you mean it is asymptotic to $\log\log n$.2012-12-22

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A more general version of this follows from Dirichlet's theorem on arithmetic progressions. In general, $\sum_{p \equiv a \pmod d} \dfrac1p \sim \dfrac1{\phi(d)}\sum_{p} \dfrac1p$ where $\phi$ is the Euler's totient function. More details can be found here, here, here and here.

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    Sorry for not responding for such a long time, I kept forgetting to respond that your answer solves original question, which can be easily deduced from standart $\delta - \epsilon$ arguments.2013-02-11