Calculus (3rd Edition)

$$\frac{4}{37}$$
Since \begin{align*} 0.108108108 &= 0.108+0.000108+0.000000108\\ &=0.108(1)+0.12(10^{-3})+ 0.12(10^{-6})+\cdots \\ &=0.108[1+\frac{1}{10^3}+\frac{1}{10^6}+\cdots ] \end{align*} Since $$1+\frac{1}{10^3}+\frac{1}{10^6}+\cdots$$ which is a geometric series and has the sum $$\frac{1}{1-\frac{1}{1000}}= \frac{1000}{999}$$ Hence $$0.108108108= 0.108 \frac{1000}{999}= \frac{108}{999} =\frac{4}{37}$$