sum of an infinite series, and extension of composite numbers to powers.

Question

Consider all the positive numbers that can be expressed as a proper power of two integers (so that neither is 1). i.e. $2^2$, $2^3$, $3^2$, $2^4$, $5^2$... and so on. And let $c$ run over all of these numbers, $4, 8, 9, 16, 25\dots$ Assuming we don't count copies as different numbers (for example $2^4 = 4^2 = 16$ so we count it once), I proved that: $\sum\dfrac1{c-1} = 1$. I was wondering if someone knew of a way to evaluate: $\sum\dfrac1c$.

Answer 1

Let $c(n)$ be the $n^\text{th}$ perfect power (without repetitions), so that $c(1)=4$, $c(2)=8$, etc. Then the sequence is basically A001597 at the OEIS.

OEIS mentions that Goldbach showed that ${\displaystyle\sum_{n=1}^\infty}\dfrac{1}{c(n)-1}=1$. And if we allow duplicates, MathWorld claims that it was also Goldbach who showed ${\displaystyle\sum_{m=2}^\infty}{\displaystyle\sum_{k=2}^\infty}\dfrac{1}{m^k}=1$.

To answer the question: OEIS and MathWorld agree that ${\displaystyle\sum_{n=1}^\infty}\dfrac{1}{c(n)}={\displaystyle\sum_{k=2}^\infty}\mu(k)(1-\zeta(k))\approx \boxed{0.87446}$ where $\mu$ is the Möbius function and $\zeta$ is the Riemann zeta function.

As an aside, OEIS claims that ${\displaystyle\sum_{n=1}^\infty}\dfrac{1}{c(n)+1}=\dfrac{\pi^2}{3}-\dfrac52$.