smc

The series can be written as the following: $\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+...$ $+\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}+...$ $+\frac{1}{10^3}+\frac{1}{10^4}+\frac{1}{10^5}+...$ and so on. by using the formula for infinite geometric series $(\frac{a}{1-r})$, We can get $\frac{\frac{1}{10}}{1-\frac{1}{10}}$ $+$ $\frac{\frac{1}{10^2}}{1-\frac{1}{10}}$ $+$ $\frac{\frac{1}{10^3}}{1-\frac{1}{10}}$ $+$ ... Since they all have common denominators, we get $\frac{(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3})}{\frac{9}{10}}$. Using the infinite series formula again, we get $\frac{\frac{\frac{1}{10}}{1-\frac{1}{10}}}{\frac{9}{10}}$ $=$ $\frac{\frac{\frac{1}{10}}{\frac{9}{10}}}{\frac{9}{10}}$ $=$ $\frac{\frac{1}{9}}{\frac{9}{10}}$ $=$ $\boxed{\frac{10}{81}}$