If `X=1+a+a^(2)+a^(3)+"..."+infty " and " y=1+b+b^(2)+b^(3)+"..."+infty "` show that `1+ab+a^(2)b^(2)+a^(3)b^(3)+"..."+infty=(xy)/(x+y-1), "where " 0ltalt1" and "0ltblt1`.

Explanation

Given, `x=1+a+a^(2)+a^(3)+"..."+infty=(1)/(1-ab)`
` implies x-ax=1`
` therefore a=((x-1))/(x)" " "......(i)"`
and `y=1+b+b^(2)+b^(3)+"..."+infty`
Similarly, ` b=((y-1))/(y)" " "......(ii)"`
Since, `0ltalt1,0ltblt1`
`therefore 0ltablt1`
Now, `1+ab+a^(2)b^(2)+a^(3)b^(3)+"..."+infty=(1)/(1-ab)`
`=(1)/(1-((x-1)/(x))((y-1)/(y)))" " ["from Eqs. (i) and (ii)"]`
`=(xy)/(xy-xy+x+y-1)`
Hence, `1+ab+a^(2)b^(2)+a^(3)b^(3)+"..."+infty =(xy)/(x+y-1)`