$35$ Prove Goldbach's theorem

$1=\frac{1}{3}+\frac{1}{7}+\frac{1}{8}+\frac{1}{15}+\frac{1}{24}+\frac{1}{26}+\frac{1}{31}+\frac{1}{35}+$dots$={\displaystyle \underset{\text{k}\text{i}\text{n}\text{P}}{\overset{?}{}}}\frac{1}{k-1}$

where P is the set of "perfect powers" defined recursively as follows:

$P=\{m^n|m2,n2,m!$inP$\}$

Perfect power

corrupts perfectly.