Question

1).  $9^{1/3} . 9^{1/9}. 9^{1/27} ........................to \ \to \infty}=?$

Answer 1

The exponents are in geometrical progression.  as the ratio is 1/3 < 1, the sum converges and we find the sum easily with the formula.

$9^{\frac{1}{3}}*9^{\frac{1}{9}}*9^{\frac{1}{27}} * 9^{\frac{1}{81}} .. .. \\ \\9^{\frac{1}{3} + \frac{1}{9} +\frac{1}{27}+\frac{1}{81}+...}\\\\=9^{\frac{1}{3}* \frac{1-\frac{1}{3^n}}{1-\frac{1}{3}}}, \ \ \ \ \frac{1}{3^n}=0,\ as\ ntends\ to\ Infinity\\ \\=9^{\frac{1}{3}*\frac{1-0}{\frac{2}{3}}}= 9^{\frac{1}{2}}\\\\x=3$