Let $\mathbb{P}$ denote the set of prime numbers. Then prove that the series $\displaystyle \sum_{\mathbb{P}}\frac{1}{p \ln p} $ converges, where $ p \in \mathbb{P}$.
Solution:
It is known that $ p \sim n \ln n $ (immediate consequence of the Prime Number Theorem) hence for large $n$ , it holds that $p_n \leq \dfrac{1}{2} n \ln n$ .
Obviously, it also holds that $\displaystyle \ln p_n \geq \ln n \Rightarrow \frac{1}{p_n \ln p_n}\leq \frac{2}{n \ln^2 n }$.
Hence from the comparison test the series converges.
Solution:
It is known that $ p \sim n \ln n $ (immediate consequence of the Prime Number Theorem) hence for large $n$ , it holds that $p_n \leq \dfrac{1}{2} n \ln n$ .
Obviously, it also holds that $\displaystyle \ln p_n \geq \ln n \Rightarrow \frac{1}{p_n \ln p_n}\leq \frac{2}{n \ln^2 n }$.
Hence from the comparison test the series converges.
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