Calculate $\sum_{n \in A} 2^{-n}$

Question

Calculate $\sum_{n \in A} 2^{-n}$ where $$A = \Big\{ n \in \mathbb N : \big| \big\{ i \in \{ 2,3,5,7\} : i \mid n \big\} \big| = 2 \Big\}.$$

Observation:
$$ x \in A \rightarrow x = p_1^{a_1} p_2^{a_2}R $$ where $$ p_1 \neq p_2 \wedge p_1, p_2 \in \left\{ 2,3,5,7\right\} \wedge \forall y \in \left\{ 2,3,5,7\right\} y \mbox{ is not divisor of } R$$

In my opinion this sum diverges: but I am not sure how can I prove that.

Answer 1

We can approach with something similar to inclusion-exclusion.

Let $2\Bbb N,3\Bbb N,5\Bbb N,7\Bbb N$ be the sets of natural numbers divisible by $2,3,5,7$ respectively.

For ease of notation, let us let $s(E,F,\dots)$ be the summation over $E\cap F\cap\dots$, so for example $s(2\Bbb N, 3\Bbb N)$ would be $\sum\limits_{n\in 2\Bbb N\cap 3\Bbb N}2^{-n}$.

We can see that to get the total sum over the elements which are in exactly two of the sets, we will have a good start by looking at $s(2\Bbb N,3\Bbb N)+s(2\Bbb N,5\Bbb N)+\dots+s(5\Bbb N,7\Bbb N)$ but we will have included things we didn't intend in doing so., having also included elements in our sum which belong to three or more sets., and having done so several times. Correcting our count, then correcting our count again, we get a final sum of:

$s(2\Bbb N,3\Bbb N)+s(2\Bbb N,5\Bbb N)+\dots+s(5\Bbb N,7\Bbb N)-3s(2\Bbb N,3\Bbb N,5\Bbb N)-\dots-3s(3\Bbb N,5\Bbb N,7\Bbb N)+6s(2\Bbb N,3\Bbb N,5\Bbb N,7\Bbb N)$

Now, recognize that for coprime $a,b,\cdots$ you have $s(a\Bbb N,b\Bbb N,\dots) = \sum\limits_{n=1}^\infty 2^{-nab\cdots} = \sum\limits_{n=1}^\infty (2^{ab\cdots})^{-n}=\dfrac{1}{2^{ab\cdots}-1}$

We get then our sum as being:

$$\dfrac{1}{2^6-1}+\dfrac{1}{2^{10}-1}+\dfrac{1}{2^{14}-1}+\dfrac{1}{2^{15}-1}+\dfrac{1}{2^{21}-1}+\dfrac{1}{2^{35}-1} - \dfrac{3}{2^{30}-1}-\dfrac{3}{2^{42}-1}-\dfrac{3}{2^{70}-1}-\dfrac{3}{2^{105}-1}+\dfrac{6}{2^{210}-1}$$

The result is $\approx 0.01694256444264848\cdots$

Answer 2

This can be broken up into 56 distinct geometric sums and summed. I used Excel and the MOD function to find all solutions up to $210 = 2\times 3\times 5\times 7$ where $\text{ROW()} \pmod{2}, \text{ROW()} \pmod{3}, \text{ROW()} \pmod{5}, \text{ROW()} \pmod{7}$ is zero exactly twice. This gives congruences mod 210 that work. So, here is the solution:

$$\left(2^{-6}+2^{-10}+2^{-12}+2^{-14}+2^{-15}+2^{-18}+2^{-20}+2^{-21}+2^{-24}+2^{-28}+2^{-35}+2^{-36}+2^{-40}+2^{-45}+2^{-48}+2^{-50}+2^{-54}+2^{-56}+2^{-63}+2^{-66}+2^{-72}+2^{-75}+2^{-78}+2^{-80}+2^{-96}+2^{-98}+2^{-100}+2^{-102}+2^{-108}+2^{-110}+2^{-112}+2^{-114}+2^{-130}+2^{-132}+2^{-135}+2^{-138}+2^{-144}+2^{-147}+2^{-154}+2^{-156}+2^{-160}+2^{-162}+2^{-165}+2^{-170}+2^{-174}+2^{-175}+2^{-182}+2^{-186}+2^{-189}+2^{-190}+2^{-192}+2^{-195}+2^{-196}+2^{-198}+2^{-200}+2^{-204}\right) \sum_{n\ge 0} 2^{-210n}$$

Where $$\sum_{n\ge 0} 2^{-210n} \approx 1$$

The result is about $0.01694256444264848$.