Let $x = 1 + a + a^{2} + . . . . .$ and $y = 1 + b + b^{2} . . . .$ where $| a | < 1$ and $| b | < 1$ . Prove that $1 + a b + a^{2} b^{2} + . . . . = \frac{x y}{x + y - 1}$

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Solution

Using $C, P$ , we get
$x = 1 + a + a^{2} + . . . . . .$
$x = \frac{1}{1 - a}$
$y = 1 + b + b^{2} + . . . . . .$
$y = \frac{1}{1 - b}$
and Let $P = 1 + a b + a^{2} b^{2} + . . . . . .$
$P = \frac{1}{1 - a b}$
and $\frac{x y}{x + y - 1} = \frac{\frac{1}{1 - a} \times \frac{1}{1 - b}}{\frac{1}{1 - a} + \frac{1}{1 - b} - 1}$
$\frac{x y}{x + y - 1} = \frac{1}{1 - b + 1 - a - 1 + a + b - a b} = \frac{1}{1 - a b}$
Since $P = \frac{1}{1 - a b} = \frac{x y}{x + y - 1}$
Hence Proved.

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If $X = 1 + a + a^{2} + . . . . \infty$ , where |a| < 1 and $y = 1 + b + b^{2} + . . . . . \infty$ where |b| <1,

prove that $1 + a b + a^{2} b^{2} + . . . . . \infty \frac{x y}{x + y - 1}$

x = \frac{a}{a + b}

y = \frac{b}{a - b}

\frac{1}{x} + \frac{1}{y} = \frac{a^{2} + b^{2}}{a b}

x = 1 + a + a^{2} + a^{3} + . . . . . \infty

y = 1 + b + b^{2} + b^{3} + . . . . . \infty

1 + a b + a^{2} b^{2} + . . . . \infty = \frac{x y}{x + y - 1}

x = 1 + a + a^{2} + a^{3} + . . . .

\infty (| a | < 1)

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

\infty (| b | < 1)

1 + a b + a^{2} b^{2} + a^{3} b^{3} + . . .

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x = 1 + a + a^{2} + a^{3} + . . .

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

1 + a b + a^{2} b^{2} + a^{3} b^{3} + . . . = \frac{x y}{x + y - 1}

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