Does sum of the reciprocals of all the composite numbers converge? 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. 
 A: 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. 
A: Simple answer : For all prime $p$ (except for 2 and 3), $p-1$ is composite,
so $\sum_{p\in\mathcal P-\{2,3\}}\frac 1 p < \sum_{p\in\mathcal P-\{2,3\}}\frac 1 {p-1} < \sum_{n\notin\mathcal P}\frac 1 n$
A: 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.
A: This sum does not converge.  Consider for example the sum of the reciprocals of all the even numbers greater than 2: $1/4 + 1/6 + 1/8 + 1/10 + \ldots$.  This diverges, the proof being analogous to the proof that the harmonic series diverges.  Thus your series has arbitrarily large partial sums and thus diverges.
