Convergence of prime series 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 $\alpha$?
 A: 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 $\alpha>0$.
A: 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 $.
A: 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 $\alpha>0$.
