My question is related to a problem in asymptotic analysis. The answer is given, but I cannot verify it.
Given the fraction $$ \frac{-\left(x+y\right)}{a+ib\left(x+y\right)^2},\quad i=\sqrt{-1},\quad a,b,x,y\in\mathbb{R}. $$ Then it holds that $$ \frac{-\left(x+y\right)}{a+ib\left(x+y\right)^2}=\frac{1}{ibx}\sum^\infty_{k=0}\left(-\frac{y}{x}\right)^k\sum^\infty_{k=0}\left(\frac{ia}{2x^2b}\right)^k=-\frac{1}{ibx}+\frac{O\left(1\right)}{x^3}. $$ I tried to crack the problem by first rewriting the fraction as $$ \frac{i}{b\left(x+y\right)\left[1-\dfrac{ia}{b\left(x+y\right)^2}\right]}=\frac{i}{b\left(x+y\right)}\cdot \frac{1}{1-\dfrac{ia}{b\left(x+y\right)^2}} $$ For the first term in the product I get (on expanding in $y$) $$ \frac{i}{b\left(x+y\right)}=-\frac{1}{ibx}\sum_{k=0}^\infty\left(-\frac{y}{x}\right)^k. $$ This would verify the first series in the answer above (without the minus sign, however). But I am lost with the remaining part. Any help is appreciated.
The statement is false. If we take $x=b=1$, $a=y=\frac12$, then the assertion is that $$\frac{-3/2}{1/2+i(3/2)^2} =\frac1i\sum_{k=0}^\infty\left(-\frac12\right)^k\sum_{k=0}^\infty\left(\frac{i}4\right)^k$$ and it is easy to verify that this isn't so.
$$\frac{-3/2}{1/2+i(3/2)^2} =\frac1i\sum_{k=0}^\infty\left(-\frac12\right)^k\sum_{k=0}^\infty\left(\frac{i}4\right)^k\implies\\ \frac{-3/2}{1/2+9i/4}=-i\frac1{1+1/2}\frac1{1-i/4}\implies \\ \frac{-6}{2+9i}=\frac{-2i}3\frac4{4-i}\implies\\ 6(12-3i)=8i(2+9i)\implies\\ 72-18i=-72+16i $$ I haven't been able to come close to the formula given.