Question: Surprised that nowhere in web or in MSE is asked it and [here](https://math.stackexchange.com/questions/1351423/does-the-sum-of-the-reciprocals-of-composites-that-are-le-1) is the not considering all of the composite numbers. 

Does $$\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+ \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \dots$$ converge? If so, how?

Both the Harmonic series and the sum of the reciprocals of all the prime numbers do not converge which doesn't help to answer the question.

Answer: 
Since $\sum\_{n=1}^\infty \frac{1}{n}$ diverges, so does $\sum\_{n=2}^\infty\frac{1}{2n}$. All the terms in this series are included in your series, so your series also diverges.
Simple answer : For all prime $p$ (except for 2 and 3), $p-1$ is composite,
so $\sum\_{p\in\mathcal P-\{2,3\}}\frac 1 p < \sum\_{p\in\mathcal P-\{2,3\}}\frac 1 {p-1} < \sum\_{n\notin\mathcal P}\frac 1 n$