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If x=1+a+a^2+oo,w h e r e|a|<1a n dy=1+b...

If `x=1+a+a^2+oo,w h e r e|a|<1a n dy=1+b+b^2+oo,w h e r e|b|<1.` prove that: `1+a b+a^2b^2+oo=(x y)/(x+y-1)`

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`x=1+a+a^2+......`
we know that sum of infinity is `a/(1-r)` so,
`x=1/(1−a)​`
Similarly, `y=1+b+b^2+......`
So infinite sum of y is `y=1/(1−b)​`
...
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