Geometric series and repeating decimals

1. First: $3\cdot\frac1{1-\frac14}=4$. Second: the series starts at $n=1$, so the sum is $\frac{q}{1-q}$ with $q=-\frac23$:
$$\frac{-2/3}{1+2/3}=-\frac25.$$

2. $0.\overline{142857}=142857\left(10^{-6}+10^{-12}+\dots\right)=142857\cdot\frac{10^{-6}}{1-10^{-6}}=\frac{142857}{999999}=\frac17.$

3. $0.999\dots=9\sum_{n\ge1}10^{-n}=9\cdot\frac{1/10}{1-1/10}=1$. The number $0.999\dots$ is *by definition* the limit of the partial sums $0.9,0.99,0.999,\dots$, and this limit is exactly $1$.