Question

x=1+a+a^2+....infinity (|a|<1)
y=1+b+b^2+....infinity (|b|<1)
Find 1+ab+a^2b^2+.....infinity

Answer 1

\[x=\frac{ 1 }{ 1-a },1-a=\frac{ 1 }{ x },a=1-\frac{ 1 }{ x }\]
similarly \[b=1-\frac{ 1 }{ y }\]
\[\left| a b \right|=\left| a \right|\left| b \right|<1\]
\[1+ab+a^2b^2+a^3b^3+...\infty=\frac{ 1 }{ 1-ab }\]
substitute the values of a and b and simplify.

Answer 2

Beautiful, but I would like to add some more:
replace ab = (1-1/x)(1-1/y) = 1-1/y -1/x+1/xy
therefore, 1-ab = 1/y +1/x-1/xy
so that \(1+ab+a^2b^2+\cdots =\dfrac{xy}{x+y-xy}\)

Answer 3

Ok...thanku both of u