Calculus: Early Transcendentals (2nd Edition)

First four terms: $0.3, 0.03, 0.003, 0.0003$ Sum: $S_\infty = \frac{1}{3}$.
The next term is the previous term multiplied by a factor of $0.1$. The infinite series is a geometric series, with a common ratio $r$ of $0.1$, which is less than $1$. Given an initial term of $a_0$ and common ratio $r$, we can find the sum by $S_\infty = \frac{a_0}{1-r}$. $S_\infty = \frac{0.3}{1-0.1} = \frac{1}{3}$.