## Calculus (3rd Edition)

$$\frac{4}{33}$$
Since \begin{align*} 0.121212 &= 0.12+0.0012+0.000012\\ &=0.12(1)+0.12(10^{-2})+ 0.12(10^{-4})+\cdots \\ &=0.12[1+\frac{1}{10^2}+\frac{1}{10^4}+\cdots ] \end{align*} Since $$1+\frac{1}{10^2}+\frac{1}{10^4}+\cdots$$ which is a geometric series and has the sum $$\frac{1}{1-\frac{1}{100}}= \frac{100}{99}$$ Hence $$0.12121212= 0.12 \frac{100}{99}= \frac{12}{99} =\frac{4}{33}$$