Sum of reciprocals of the perfect powers 
*

*What is the sum of all of the numbers which can be written as the reciprocal of an integer that is also a power greater than one, excluding the powers of one ?.

*I'm asking about the sum of numbers which can be written as powers. $$
\mbox{This sum would start off as}\quad
\frac{1}{4} + \frac{1}{8} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \cdots
$$


Note that I didn't count $1/16$ twice, as we are only adding numbers which can be written in that form.
-I can see how this alternative sum would have probably been easier to handle, since we can see that it equates to the sum $\zeta\left(s\right) - 1$ for all $s$ larger than one.
 - I know that the first sum must converge, because I have worked out that the sum of numbers which can be written in the form $1/\left(a^{b} - 1\right)$ converges to $1$, where
$a,b > 1$.
Is there any way of evaluating the first sum ?. Also, does the second sum converge as well and how can we evaluate it ?.
I have made a program $\left(~\texttt{python}~\right)$ which carries out partial sums for these sorts of series. I'll attach the source code if anyone wants it.

Your sum contains the sum of all reciprocals of non-unit natural number powers of primes, that is, every number of the form $\frac{1}{p^k}$ for prime $p$, and $k\in \{2,3,...\}=\mathbb{N}\setminus\{1\}$. Since each number of this form is uniquely determined by the choice of $p$ and $k$, we obtain
$\sum_{p}\sum_{i=2}^{\infty}\dfrac{1}{p^i}=\sum_p\dfrac{1}{p(p-1)}$ where $p$ is every prime.
Since $\sum_{i=1}^{\infty}\dfrac{1}{p^i}=\dfrac{1}{p-1}$ (this follows from some basic number theory; an easy way to see it is to consider the repeating decimal, $0.111...$ in base $p$), yet we have removed the leading term, $\dfrac{1}{p}$, hence
$\sum_{p}\sum_{i=2}^{\infty}\dfrac{1}{p^i}=\dfrac{1}{p-1}-\dfrac{1}{p}=\dfrac{p}{p(p-1)}-\dfrac{p-1}{p(p-1)}=\dfrac{p-(p-1)}{p(p-1)}=\dfrac{1}{p(p-1)}$
As for the case at hand, we are no longer concentrating on primes, but the prior counting result holds for non-prime numbers, $n$, the problem now is: how to avoid double counting numbers such as 4^2=4^2=16? Again the solution is to pick on primes, the prime here is $2$, thus we will count that as the $2^4$ and this tactic, in fact, handles all powers of $4$. But that doesn't solve the issue raised by numbers such as $36=2^2\cdot 3^2$. Here we have distinct primes that are part of the factorization, in this case, we can count this as $6^2$. We also have to deal with cases such as
$4^2\cdot 3^2=2^4\cdot 3^2=12^2=144$
For this, we consider it as $12^2$. By now the pattern is maybe emerging: We want take our sum of (non-unit) reciprocal powers over every number whose exponents of its prime factorization have gcd of 1. That is, We are interested in numbers:
$b=p_1^{k_1}\cdot p_2^{k_2}\cdot ...\cdot p_f^{k_f}$
for which $\gcd(k_1,...,k_f)=1$. The reason why this is the criterion we require is that we seek to eliminate recounts of numbers we already have, and any number which has the above $\gcd(k_1,...,k_f)=r\neq 1$, then we can take the $r$th root of that number, and obtain a number which does have $\gcd(k_1,...,k_f)=1$, hence we have a unique way of representing every number we care about as a power of a number in the above form with $\gcd(k_1,...,k_f)=1$.
Thus our sum is, in terms of the above notation:
$\sum_b\sum_{i=2}^{\infty}\dfrac{1}{b^i}=\sum_b\dfrac{1}{b(b-1)}$.
I am not sure if there is a nice way of characterizing the relevant numbers $b$ or not, but provided a suitably nice characterization can be found this may prove useful.
