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Question Number 45283 by peter frank last updated on 11/Oct/18

$$\mathbf{if}\mathbf{x}=1+\mathbf{a}+{\mathbf{a}}^2+{\mathbf{a}}^3+\dots \dots \dots$$$$\mathbf{y}=1+\mathbf{b}+{\mathbf{b}}^2+{\mathbf{b}}^3+\dots \dots \dots$$$$\mathbf{prove}\mathbf{that}$$$$1+\mathbf{ab}+{\mathbf{a}}^2{\mathbf{b}}^2+{\mathbf{a}}^3{\mathbf{b}}^3+\dots \dots \dots =\frac{\mathbf{xy}}{\mathbf{x}+\mathbf{y}-1}$$

Answered by ajfour last updated on 11/Oct/18

$$x=\frac{1}{1-a};y=\frac{1}{1-b}$$$$z=1+ab+a^2{b}^2+a^3{b}^3+...$$$$=\frac{1}{1-ab}=\frac{1}{1-(1-\frac{1}{x})(1-\frac{1}{y})}$$$$=\frac{xy}{xy-(x-1)(y-1)}$$$$=\frac{xy}{x+y-1}.$$

Commented by peter frank last updated on 11/Oct/18

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by maxmathsup by imad last updated on 11/Oct/18

$$thisistrueifweknowthat\mid a\mid <1and\mid b\mid <1...$$

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