Prove that: (9^1/3 . 9^1/9 . 9^1/27 ... ∞) = 3.

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Solution

$LHS = 9^{\frac{1}{3}} . 9^{\frac{1}{9}} . 9^{\frac{1}{27} . . . \infty} = 9^{(\frac{1}{3} + \frac{1}{9} \frac{1}{27} .)} = 9^{\{\frac{(\frac{1}{3})}{(1 - \frac{1}{3})}\}} = 9^{\frac{(\frac{1}{3})}{(1 - \frac{1}{3})}} = \sqrt{9} = 3 = RHS$

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9^{\frac{1}{3}} {.9}^{\frac{1}{9}} {.9}^{\frac{1}{27}} . . . . . . . . . . .

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