# GP Series Problem...stuck

• Apr 29th 2009, 08:36 AM
chakravarthiponmudi
GP Series Problem...stuck
I am trying to solve the problem and but i am stuck.. Can someone please have look and help out...
Problem
if $\displaystyle x=1+a+a^2+...$ and $\displaystyle y=1+b+b^2$ where |a| <1 and |b| < 1. Prove that $\displaystyle 1 + ab + a^2b^2 + ...= \frac{xy}{x+y-1}$

Solution
$\displaystyle x=1+a+a^2+...$
GP Series
$\displaystyle x = \frac{1-a^n}{1-a} [1]$

Similarly
$\displaystyle y=\frac{1-b^n}{1-b} [2]$

taking LHS of the equation
$\displaystyle 1 + ab + a^2b^2 + ...$
using GP
$\displaystyle \frac{1-(ab)^n}{1-ab} [3]$

RHS of the equation
$\displaystyle \frac{xy}{x+y-1} [4]$

If I substitute 1 and 2 in 4,i get

$\displaystyle \frac{(1-a^n)(1-b^n)}{(1-a^n)(1-b)+(1-b^n)(1-a)-(1-a)(1-b)}$

this is no where near to equ 4. My approach might be wrong. please suggest if possible...
• Apr 29th 2009, 09:12 AM
Media_Man
You're so close...
Quote:

$\displaystyle \frac{1-(ab)^n}{1-ab}$
Quote:

$\displaystyle \frac{(1-a^n)(1-b^n)}{(1-a^n)(1-b)+(1-b^n)(1-a)-(1-a)(1-b)}$
You're so close! Now, find the limit of both sides as n goes to infinity. Since |a|<1 and |b|<1, $\displaystyle a^n$, $\displaystyle b^n$, and $\displaystyle (ab)^n$ go to zero. Go ahead and make these substitutions and both the LHS and RHS will agree!

*In the original problem, are you required to prove that this equality holds for a terminating sequence $\displaystyle x=1+a+a^2+...+a^n$ ??? If infinite, then your equations should read:

$\displaystyle x = \frac{1}{1-a} [1]$

$\displaystyle y = \frac{1}{1-b} [2]$
• Apr 29th 2009, 09:22 AM
chakravarthiponmudi
Thank you.
No they are not terminating sequence... They are infinite sequence.. I assumed it wrongly... Thanks for correcting me...