Binomial expansion and convergence

Question

May I ask for some help on this?

Answer 1

Hint

$$(2x-5y)^{-1}=\frac 1{2x-5y}=\frac 1{2x(1-\frac{5y}{2x})}=\frac 1{2x}\times\frac 1{1-\frac{5y}{2x}}$$ Take care that the above steps assume $x\neq 0$.

Now, remember the expansion $$\frac 1{1-t}=1+t+t^2+\cdots+t^n+\cdots$$ which is valid if $|t|<1$. Replace $t$ by $\frac{5y}{2x}$.

I am sure that you can take it from here.