help in series

**arangu1508** · November 18th 2012, 10:18 PM

If x = 1 + a + a^2 + a^3 + ...

y = 1 + b + b^2 + b^3 + ...

then show that 1 + ab + a^2b^2 + a^3b^3 + ... = xy/(x + y - 1).

How to go about this. Guidance required.

Thanks.

Aranga

**Prove It** · November 18th 2012, 10:31 PM

Originally Posted by arangu1508

If x = 1 + a + a^2 + a^3 + ...

y = 1 + b + b^2 + b^3 + ...

then show that 1 + ab + a^2b^2 + a^3b^3 + ... = xy/(x + y - 1).

How to go about this. Guidance required.

Thanks.

Aranga

I'm going to assume that $\displaystyle \begin{align*} |a| < 1 \end{align*}$ and $\displaystyle \begin{align*} |b| < 1 \end{align*}$ . Then

$\displaystyle \begin{align*} x &= 1 + a + a^2 + a^3 + \dots \\ &= \frac{1}{1 - a} \\ \\ y &= 1 + b + b^2 + b^3 + \dots \\ &= \frac{1}{1 - b} \\ \\ \frac{x \, y}{x + y - 1} &= \frac{\left( \frac{1}{1 - a} \right) \left( \frac{1}{1 - b} \right) }{ \frac{1}{1 - a} + \frac{1}{1 - b} - 1 } \\ &= \frac{\frac{1}{(1 - a)(1 - b)}}{\frac{1 - b + 1 - a - (1 - b)(1 - a) }{(1 - a)(1 - b)}} \\ &= \frac{1}{1 - b + 1 - a - (1 - b)(1 - a)} \\ &= \frac{1}{2 - a - b - (1 - a - b + ab)} \\ &= \frac{1}{1 - ab} \\ &= 1 + ab + (ab)^2 + (ab)^3 + \dots \\ &= 1 + ab + a^2b^2 + a^3b^3 + \dots \end{align*}$

**MarkFL** · November 18th 2012, 10:45 PM

We are given:

$x=1+a+a^2+a^3+\cdots$

Multiply through by $a$ :

$ax=a+a^3+a^4+\cdots=x-1$

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

Likewise, we find:

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

Now, use a similar method, where:

$f(x,y)=1+ab+(ab)^2+(ab)^3+\cdots$

Then, after multiplying through by $ab$ solve for $f(x,y)$ , then use your values of $a$ and $b$ to get the desired result.

**Soroban** · November 19th 2012, 06:40 AM

Hello, arangu1508!

Another approach . . .

$\text{If }\,\begin{Bmatrix}x \:=\: 1 + a + a^2 + a^3 + \hdots & [1] \\ y \:=\: 1 + b + b^2 + b^3 + \hdots & [2] \end{Bmatrix}$

$\text{then show that: }\:S \;=\;1 + ab + a^2b^2 + a^3b^3 + \cdots \:=\: \frac{xy}{x + y - 1}$

From [1]: . $x \:=\:\frac{1}{1-a} \quad\Rightarrow\quad a \:=\:\frac{x-1}{x}\;\;[3]$

From [2]: . $y \:=\:\frac{1}{1-b} \quad\Rightarrow\quad b \:=\:\frac{y-1}{y}\;\;[4]$

We have: . $S \;=\;1 + (ab) + (ab)^2 + (ab)^3 + \cdots \;=\;\frac{1}{1-ab}\;\;[5]$

$\text{Substitute [3] and [4] into [5]: }\:S \;=\;\frac{1}{1-\left(\frac{x-1}{x}\right)\left(\frac{y-1}{y}\right)} \;=\;\frac{1}{1 - \frac{(x-1)(y-1)}{xy}}$

$\text{Multiply by }\frac{xy}{xy}\!:\;S \;=\;\frac{xy}{xy - (x-1)(y-1)} \;=\; \frac{xy}{xy - xy + x + y - 1}$

$\text{Therefore: }\:S \;=\;\frac{xy}{x+y-1}$

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