Convergence of $\sum_{m\text{ is composite}}\frac{1}{m}$ It can be easily show that the harmonic series $$\sum_{n=1}^{\infty}\dfrac{1}{n}$$ is divergent.
Also it has shown that the infinite series of reciprocals of primes $$\sum_{p\text{ is prime}}\dfrac{1}{p}$$ is divergent.  

I believe that the series $$\sum_{m\text{ is composite}}\dfrac{1}{m}$$ is also divergent.

But I have no idea to attempt for a proof. Any help will be appreciate. Thank you. 
 A: For any prime $p$, the number $2p$ is composite. Therefore
$$
\sum_{m \text{ composite}} \frac{1}{m} \geq \sum_{p \text{ prime}} \frac{1}{2p} = \frac12 \sum_{p \text{ prime}} \frac{1}{p}.
$$
Since the latter series is divergent, the same holds for the first series.
A: $$\sum_{m\text{ is composite}}\frac{1}{m}>\sum_{\substack{m \text{ is even}\\m>2}}\frac{1}{m}$$
and the latter is clearly divergent. 
If you like to nuke it, by the Mertens' second theorem it follows that:
$$\sum_{\substack{m\text{ is composite}\\m\leq M}}\frac{1}{m}=\log M-\log\log M+O(1).$$
A: Let's order the primes $p_n$ being the $n^{th}$ prime of the set of primes. For each composite number $m$ there is a prime such that $p_{n-1}\lt m\lt p_n$ and therefore $\frac{1}{m}\gt \frac{1}{p_n}$ if $\alpha_n$ is the number of composites between two consecutive primes, one has $\alpha_n\geq 2$ and we can write
$$\sum_{m\text{ is composite}}\frac{1}{m}\gt \sum_{n}\alpha_n\frac{1}{p_n}\geq\sum_{n}2\frac{1}{p_n}$$
and this tells us that the sum is divergent
A: $$[2(2),2(3),2(5)\cdots]\in[4,6,8,9,10\cdots]$$
