Silk Road Mathematics Competition

Answer: the sum is equal $1$. This can be seen as follows: $$\sum _{k\in A}\frac{1}{k-1}=\sum _{k\in A}\left(\frac{1}{k}\cdot \frac{1}{1-1/k}\right)=\sum _{k\in A}\sum _{i\ge 1}\frac{1}{k^i}$$ Proposition: For $x\in \mathbb{Z}$ the sets $\{(m,n):x=m^n;m,n\in \mathbb{Z},m,n\ge 2\}$ and $\{(k,i):x=k^i;k\in A,i\in \mathbb{N}\}$ have the same number of elements. Proof of the proposition: Let $x=p_1^{a_1}\dots p_t^{a_t}$ be the prime factorization of $x$. $a=\text{GCD}(a_1,\dots ,a_t)$, $a_j=a{b}_j$ for $0\le j\le t$ and $y=p_1^{b_1}\dots p_t^{b_t}$. Then $x=m^n$ iff $n\mid a$ and $m=y^{a/n}$. So both of the sets above have $\tau (a)-1$ elements, where $\tau (a)$ stands for the number of positive divisors of $a$. $\square$

Therefore: $$\begin{array}{rl}\sum _{k\in A}\sum _{i\ge 1}\frac{1}{k^i}& =\sum _{m\ge 2}\sum _{n\ge 2}\frac{1}{m^n}\\ & =\sum _{m\ge 2}\frac{1/m^2}{1-1/m}=\sum _{m\ge 2}\frac{1}{m(m-1)}\end{array}$$ And finally, $$\sum _{m=2}^l\frac{1}{m(m-1)}=\sum _{m=2}^l\left(\frac{1}{m-1}-\frac{1}{m}\right)=1-\frac{1}{l}\to 1\text{ for }l\to \infty .$$