Question

The sum of : $\frac{1}{9^3}.\frac{1}{9^9}.\frac{1}{9^{27}}......\infty$ is

Answer 1

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

$=9^{-3}{.9}^{-9}{.9}^{-27}......$

$=9^{-3-9-\mathrm{27......}}$

$=9^{-3\left(1+3+9+27+.....\right)}$

$=9^{-3\times \frac{1}{1-3}}$

$=9^{-3\times -\frac{1}{2}}$

$=9^{\frac{3}{2}}$

$={\left(3\right)}^3$

$=27$