Does sum of the reciprocals of all the composite numbers converge? [duplicate]

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.

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 $\sum_{p\in\mathcal P-\{2,3\}}\frac 1 p < \sum_{p\in\mathcal P-\{2,3\}}\frac 1 {p-1} < \sum_{n\notin\mathcal P}\frac 1 n$

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.