Question

Prove that :

$(9^{\frac{1}{3}}{.9}^{\frac{1}{9}}{.99}^{\frac{1}{27}}{.169}^{\frac{1}{32}}.....\infty )=3.$

Answer 1

$9^{\frac{1}{3}}\times 9^{\frac{1}{9}}\times 9^{\frac{1}{27}}...\infty =9^{(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\infty )}=9^{\left(\frac{\frac{1}{3}}{1-\frac{1}{3}}\right)[Using{S}_{\infty }=\frac{a}{1-r}]}=9^{(\frac{1}{3}\times \frac{3}{2})}=9^{\frac{1}{2}}=3$