TITLE: Sum of reciprocals of the perfect powers
QUESTION [15 upvotes]: 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.

REPLY [6 votes]: $$\begin{align} \sum_{n=2}^{\infty} \left [\zeta(n)-1 \right ] &=\sum_{n=2}^{\infty}  \sum_{k=2}^{\infty} \frac1{k^n}\\ &= \sum_{k=2}^{\infty}  \sum_{n=2}^{\infty} \frac1{k^n}\\ &= \sum_{k=2}^{\infty} \left ( \frac{k}{k-1} - \frac{k+1}{k}\right ) \\ &= \sum_{k=2}^{\infty} \frac1{k (k-1) }\\ &= 1\end{align}$$
NB
$$ \sum_{n=2}^{\infty} \frac1{k^n} = \sum_{n=0}^{\infty} \frac1{k^n} - 1 - \frac1{k} = \frac{k}{k-1} - \frac{k+1}{k} $$