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.
If $p\_n$ is the $n$th prime number,

Note $$(\sum\_{i=1}^{n}\frac{1}{p\_i})^2(\text{Divergent})=\sum\_{i=1}^{n}\frac{1}{(p\_i)^2}(\text{Convergent})+2\sum\_{1 \le i < j \le n}^{n}\frac{1}{p\_ip\_j}$$ Is true. 

We have that $$\sum\_{1 \le i \le j \le n}^{n}\frac{1}{p\_ip\_j}$$Diverges. But all numbers that are a multiple of two primes are composite.