I believe I have proved that, for all $q\notin[-1,1]$, it follows
$$\frac{1}{q}=\sum_{\text{prime }p}^\infty\frac{1}{q^{p-1}-1}$$
Of course this is a huge result, so I want to see if my proof is valid.
For $|q|<1$, it follows that
$$\begin{align}\frac{q}{1-q}&=q+q^2+q^3+q^4+\cdots \\ \therefore \frac{q^2}{1-q^2}&=q^2+q^4+q^6+q^8+\cdots\end{align}$$ $$\therefore \frac{q}{1-q}-\frac{q^2}{1-q^2}=q+q^3+q^5+q^7+\cdots$$ We also have that: $$\frac{q^3}{1-q^3}=q^3+q^6+q^9+q^{12}+\cdots$$ $$\therefore \frac{q}{1-q}-\frac{q^2}{1-q^2}-\frac{q^3}{1-q^3}=q+q^5+q^7+q^{11}+\cdots$$ and if we keep subtracting $\cfrac{q^p}{1-q^p}$ for the next prime $p$, we will be left with $q$. $$\therefore \frac{q}{1-q}-\sum_{\text{prime }p}^{\infty}\frac{q^p}{1-q^p}=q$$ $$\therefore \frac{1}{1-q}=\sum_{\text{prime }p}^{\infty}\frac{q^{p-2}}{1-q^p}$$ Now let $q\mapsto \cfrac 1q$. Since our original constraint was $|q|<1$, then $q\mapsto \cfrac 1q$ implies $\bigg|\cfrac 1q\bigg|<1$, or, $q\notin[-1,1]$. $$\therefore \cfrac{1}{1-\frac 1q}=\sum_{\text{prime }p}^{\infty}\frac{1}{q^{p-2}\big(1-\frac{1}{q^p}\big)}$$ for any $q\notin[-1,1]$. Multiply both sides by $\cfrac{q-1}{q^2}$ and we have: $$\cfrac 1q=\sum_{\text{prime }p}^\infty\frac{1}{q^{p-1}-1}$$
Is this a valid proof? If so, has this result been discovered before?
Thanks.
We have $\frac1{1-q^p}=\sum_{p\mid k}q^k$, so $$ \sum_{p\text{ prime}}\frac{q^{p-2}}{1-q^p}=\sum_{p\text{ prime}}\sum_{p\mid k+2}q^k=\sum_{k=0}^\infty q^k\sum_{p\mid k+2}=\sum_{k=0}^\infty\omega(k+2)q^k$$ and this is not the same as $\frac1{1-q}=\sum_{k=0}^\infty q^k$.
Also, checking the claim itself for $q>1$, already the first of all positive summands is $\frac1{q-1} > \frac1q$.