Using the geometric series ∑n=0∞rn=11−r\displaystyle \sum_{n=0}^{\infty} r^n = \frac{1}{1-r}n=0∑∞rn=1−r1 for ∣r∣<1|r| < 1∣r∣<1, what is the sum ∑n=0∞(13)n\displaystyle \sum_{n=0}^{\infty} \left(\frac{1}{3}\right)^nn=0∑∞(31)n?
13\frac{1}{3}31
32\frac{3}{2}23
12\frac{1}{2}21
Diverges