TITLE: Does sum of the reciprocals of all the composite numbers converge?
QUESTION [23 upvotes]: Surprised that nowhere in web or in MSE is asked it and here 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.

REPLY [6 votes]: 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.