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.
 A: $$\begin{align} \sum_{n=2}^{\infty} \left [\zeta(n)-1 \right ] &=\sum_{n=2}^{\infty}  \sum_{k=2}^{\infty} \frac1{k^n}\\ &= \sum_{k=2}^{\infty}  \sum_{n=2}^{\infty} \frac1{k^n}\\ &= \sum_{k=2}^{\infty} \left ( \frac{k}{k-1} - \frac{k+1}{k}\right ) \\ &= \sum_{k=2}^{\infty} \frac1{k (k-1) }\\ &= 1\end{align}$$
NB
$$ \sum_{n=2}^{\infty} \frac1{k^n} = \sum_{n=0}^{\infty} \frac1{k^n} - 1 - \frac1{k} = \frac{k}{k-1} - \frac{k+1}{k} $$
A: 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.
A: The sum of the reciprocals of the perfect powers should be
$$\sum_{k=2}^{\infty}\sum_{j=2}^{\infty}\frac{(-\mu(k))}{j^k}=
\sum_{k=2}^{\infty}\mu(k)\left(1-\zeta(k)\right)
\approx 0.874464$$
Where $\mu$ is the Mobius function.
See the OEIS sequence A001597.
In order to verify the above formula, it suffices to show that in that double sum each perfect power is counted once. Now $n>1$ is a perfect power iff its prime factorization $p_1^{a_1}\cdots p_{r}^{a_r}$ satisfies $d:=\gcd(a_1,\dots, a_r)>1$. 
In the above double sum, the denominator $j^k$ is equal to $n$ iff $k>1$ is a divisor of $d$ and $j=n^{1/k}$. Hence the fraction $1/n$ is counted the following number of times
$$\sum_{k\mid d,k>1}(-\mu(k))=0+\mu(1)=1$$
and we are done.
