Calculus: Early Transcendentals (2nd Edition)

$$\sum_{k=0}^\infty \left(\frac{1}{4} \right)^k=\frac{4}{3}$$
To determine what an infinite geometric series converges to we use the following formula: $$\sum_{k=0}^\infty ar^k=\frac{a}{1-r}$$ so we must identify $a$ and $r$ for our series $$\sum_{k=0}^\infty \left(\frac{1}{4} \right)^k$$ $a=1$ and $r=\frac{1}{4}$ we plug this into above formula: $$\sum_{k=0}^\infty \left(\frac{1}{4} \right)^k=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}$$