What does this converge to? Consider the sum of the reciprocals of all numbers of the form $p^n$ with $n ≥ 2$ and $p$ prime, this is shown to converge by noting that it is strictly less then this convergent sum https://en.wikipedia.org/wiki/Perfect_power.
Does anyone know what this sum is or where I can find information?
 A: I don't think there's a closed form for it. As others mentioned in comments, there's a connection with the prime zeta function:
$$S=\sum_{p\,prime}\sum^\infty_{n=2}\frac1{p^n}=\sum_{p\,prime}\frac1{p(p-1)}=\sum^\infty_{n=2}P(n),$$ and we can calculate the prime zeta function easily through $$P(s)=\sum^\infty_{n=1}\frac{\mu(n)}{n}\,\ln\zeta(ns),$$ where $\mu$ is Moebius function. Obviously, $P(n)=O(2^{-n}),$ so the series converges rapidly. We can use that to get a better numerical value, but we can also substitute the expression for $P(n)$ and change the order of summation:
$$S=\sum^\infty_{m=2}\left(\sum_{n|m, n<m}\frac{\mu(n)}{n}\right)\,\ln\zeta(m).$$
Now we know that $$\sum_{n|m}\frac{\mu(n)}{n}=\frac{\varphi(m)}{m},$$  where $\varphi$ is the totient function, but the terms with $m=n$ are missing, because $P(1)$ is missing in our sum, for obvious reasons. So we finally have the curious result
$$S=\sum^\infty_{m=2}\frac{\varphi(m)-\mu(m)}{m}\,\ln\zeta(m).$$ As was to be expected, both formulas give the same numerical approximation, $S=0.7731566690497961$.
