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

This question already has an answer here:

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.

marked as duplicate by Bumblebee, John B, Ethan Bolker, Alex M., Fly by NightApr 19 '16 at 16:51

• The primes have zero natural density so... – PITTALUGA Apr 8 '16 at 15:58
• You have a lower bound given by the sum of all reciprocals of even numbers bigger than 2, which is essentially half of the harmonic series. So it must diverge. – Brandon Carter Apr 8 '16 at 15:58
• math.stackexchange.com/questions/1211498/… – Bumblebee Apr 19 '16 at 6:12

Simple answer : For all prime $p$ (except for 2 and 3), $p-1$ is composite, so $\sum_{p\in\mathcal P-\{2,3\}}\frac 1 p < \sum_{p\in\mathcal P-\{2,3\}}\frac 1 {p-1} < \sum_{n\notin\mathcal P}\frac 1 n$

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.

• I like this answer way better than the accepted one - it's much more intuitive IMO. – fluffy Apr 8 '16 at 21:30
• I think it is important to mention that this property (that adding extra terms cannot make a divergent series converge) is only true for series where each term is positive. Otherwise the alternating harmonic series is an easy counterexample. – Fengyang Wang Apr 9 '16 at 4:11
• @FengyangWang: I disagree -- I think this answer is exactly as long as it needs to be. – TonyK Apr 9 '16 at 14:31
• @fluffy - This answer is very nice indeed (and +1). The accepted answer is a bit more 'romantic' ;) – user231343 Apr 9 '16 at 20:14
• I like this answer, the sum of reciprocals of primes is a big jackhammer – enthdegree Apr 14 '16 at 8:49

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.

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.