Question

Let $x=1+a+a^2+.....$ and $y=1+b+b^2....$ where $\left|a\right|<1$ and $\left|b\right|<1$. Prove that $1+ab+a^2{b}^2+....=\frac{xy}{x+y-1}$

Answer 1

Using $C,P$, we get

$x=1+a+a^2+......$

$x=\frac{1}{1-a}$

$y=1+b+b^2+......$

$y=\frac{1}{1-b}$

and Let $P=1+ab+a^2{b}^2+......$

$P=\frac{1}{1-ab}$

and $\frac{xy}{x+y-1}=\frac{\frac{1}{1-a}\times \frac{1}{1-b}}{\frac{1}{1-a}+\frac{1}{1-b}-1}$

$\frac{xy}{x+y-1}=\frac{1}{1-b+1-a-1+a+b-ab}=\frac{1}{1-ab}$

Since $P=\frac{1}{1-ab}=\frac{xy}{x+y-1}$

Hence Proved.