I have had this conjecture for a while:
Let $q\in\mathbb R\setminus [-1,1]$ and let $p_n$ be the $n^\text{th}$ prime. Then: $$\sum_{n=1}^{\infty}\bigg(\frac 1{q^{p_n}-1}-\sum_{k=1}^{n-1}\frac 1{q^{p_n p_{n-k}}-1}\bigg)=\frac 1{q(q-1)}$$
The sum looks like this when we expand it out:
$$\frac 1{q^2-1}+\bigg(\frac 1{q^3-1}-\frac 1{q^{3\times 2}-1}\bigg)+\bigg(\frac 1{q^5-1}-\frac 1{q^{5\times 3}-1}-\frac 1{q^{5\times 2}-1}\bigg)+\cdots=\frac 1{q(q-1)}$$
I don't know how to show something like this. I can pull out a factor of $1/(q-1)$ but I don't think that will help.
Is this even true? I just had it written down in my notebook. (Sorry for not having any good proof attempts of my own.)
The formula is (closely) but not exactly true. Note that
\begin{align*} \sum_{n=1}^{\infty} \biggl[ \frac{1}{q^{p_n} - 1} - \sum_{k=1}^{n-1} \frac{1}{q^{p_n p_k} - 1} \biggr] &= \sum_{n \geq 1} \sum_{d \geq 1} \frac{1}{q^{d p_n}} - \sum_{n > k \geq 1} \sum_{d \geq 1} \frac{1}{q^{d p_n p_k}} \\ &= \sum_{l \geq 1} \biggl[ \sum_{p \mid l} 1 - \sum_{\substack{p p' \mid l \\ p < p'}} 1 \biggr] \frac{1}{q^l} \\ &= \sum_{l=1}^{\infty} \biggl[ \omega(l) - \binom{\omega(l)}{2} \biggr] \frac{1}{q^l}, \end{align*}
where $\omega(l) $ counts the number of distinct prime factors of $l$. Writing $a_l = \omega(l) - \binom{\omega(l)}{2}$ and noting that $k-\binom{k}{2} = 1$ for $k = 1, 2$, we get
$$ a_1 = 0, \qquad a_2 = a_3 = \ldots = a_{29} = 1, \qquad a_{30} = 0, \qquad \ldots $$
(This has to do with the fact that $30 = 2\cdot3\cdot5$ is the first positive integer with $3$ or more distinct prime factors.) Then this gives
$$ \frac{1}{q(q-1)} - \sum_{n=1}^{\infty} \biggl[ \frac{1}{q^{p_n} - 1} - \sum_{k=1}^{n-1} \frac{1}{q^{p_n p_k} - 1} \biggr] = \sum_{l=2}^{\infty} \frac{1 - a_l}{q^l} = \frac{1}{q^{30}} + \cdots. $$
In particular, the difference is quite small if $q$ is large. For example, if $q = 3$ then
$$ \sum_{n=1}^{\infty} \biggl[ \frac{1}{3^{p_n} - 1} - \sum_{k=1}^{n-1} \frac{1}{3^{p_n p_k} - 1} \biggr] \approx 0.16666666666666180972\cdots, $$
which is quite close to but not the same as $\frac{1}{3(3-1)} = \frac{1}{6}$.