<s>[INST] How find $\sum_{k \in A} \frac{1}{k-1} $ for $ A = \{ m^n| \text{ } m, n \in Z \text { and } m, n \ge 2 \} $ If $ A = \{ m^n| \text{  } m, n \in Z  \text {  and  } m, n \ge 2 \} $, then how find $\sum_{k \in A} \frac{1}{k-1} $?
 [/INST] The answer is 1.
The result is known as Goldbach-Euler theorem.
See Wikipedia entry for "proof".
For rigorous proof, you could consider sum of reciprocals of all perfect powers, $S$. Note that sum equals
$$
S = \sum_{x \in B}\sum_{n = 2}^{\infty}\frac{1}{x^{n}} = \frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\ldots+\frac{1}{25}+\frac{1}{125}+\ldots
$$
where $B$ is set of all integers not perfect power. Why? All reciprocals of perfect powers are clearly in the sum, and terms in sum are distinct: if
$$
\frac{1}{x^{n}} = \frac{1}{y^{m}}
$$
and $x$ and $y$ are distinct, also $m$ and $n$ are. But since $x$ and $y$ aren't perfect powers, $x^{n}$ and $y^{m}$ are exactly $n$:th and $m$:th powers, respectively, for distinct $m$ and $n$ which is impossible.
But now
$$
S = \sum_{x \in B}\sum_{n = 2}^{\infty}\frac{1}{x^{n}} = \sum_{x = 2}^{\infty}\sum_{n = 2}^{\infty}\frac{1}{x^{n}}-\sum_{x \in A}\sum_{n = 2}^{\infty}\frac{1}{x^{n}} \\
= \sum_{x = 2}^{\infty}\frac{1}{x(x-1)}-\sum_{x \in A}\frac{1}{x(x-1)} \\
= 1- \sum_{x \in A}\left(\frac{1}{x-1}-\frac{1}{x}\right) = 1- \sum_{x \in A}\frac{1}{x-1}+\sum_{x \in A}\frac{1}{x} \\
= 1- \sum_{x \in A}\frac{1}{x-1}+S
$$
so
$$
\sum_{x \in A}\frac{1}{x-1} = 1
$$
</s>