2011/057) Given |a | < 1 and |b| < 1

If $1+a+a^2+\cdots =x$ and $1+b+b^2+\cdots.=y$ then find $1+ab+a^2.b^2+.... $ in terms of x and y

Solution
we have
$x = \dfrac{1}{1-a}$
or $(1-a) = \dfrac{1}{x}$
or $a = 1 – \dfrac{1}{x} = \dfrac{x-1}{x}$
similarly
$b = \dfrac{y-1}{y}$
now as |a | and |b| both < 1 so |ab| < 1 and hence
$1+ ab + a^2b^2 + a^3b^3 + \cdots= \dfrac{1}{1-ab} = \dfrac{1}{1- \frac{x-1}{x} * \frac{y-1}{y}}$
= $\dfrac{xy}{xy - (x - 1)(y - 1)}$
= $\dfrac{xy}{x + y - 1}$