<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] Not a complete answer to your question, but if $A$ is a multi-set (where any element may appear more than once), then your series is larger than the sum of all the following series put together:


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*$\sum\limits_{n=2}^{\infty}\frac{1}{2^n}=\frac{1}{2-1}-\frac{1}{2}=\frac{1}{2}$

*$\sum\limits_{n=2}^{\infty}\frac{1}{3^n}=\frac{1}{3-1}-\frac{1}{3}=\frac{1}{6}$

*$\sum\limits_{n=2}^{\infty}\frac{1}{4^n}=\frac{1}{4-1}-\frac{1}{4}=\frac{1}{12}$

*$\sum\limits_{n=2}^{\infty}\frac{1}{5^n}=\frac{1}{5-1}-\frac{1}{5}=\frac{1}{20}$

*$\dots$


In other words:
$$\sum\limits_{k\in{A}}\frac{1}{k-1}>\sum\limits_{k\in{A}}\frac{1}{k}=\sum\limits_{m=2}^{\infty}\sum\limits_{n=2}^{\infty}\frac{1}{m^n}=\sum\limits_{m=1}^{\infty}\frac{1}{m(m+1)}=1$$
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