# How do you find the sum of the infinite geometric series 3 + 1 + 1/3 + 1/9 + ...?

Note that $\frac{a}{1 - a}$ yields the infinite sum $a + {a}^{2} + {a}^{3} + {a}^{4} + \ldots$.
(One can do the specified division to see why this is true.) Substituting a=1/3, one gets that the infinite sum $\frac{1}{3} + \frac{1}{3} ^ 2 + \frac{1}{3} ^ 3 + \ldots$ is equal to $\left(\frac{1}{3} / \left(1 - \frac{1}{3}\right)\right)$ = $\frac{\frac{1}{3}}{\frac{2}{3}}$ = $\frac{1}{2}$. So, adding one to this infinite sum yields $\frac{3}{2}$. Note that the given infinite sum is three times our sum, i.e. one can factor 3 out of each term, to get $3 \left(1 + \frac{1}{3} + \frac{1}{3} ^ 2 + \frac{1}{3} ^ 3 + \ldots\right)$, so the sum of the given series is $3 \left(\frac{3}{2}\right)$ =$\frac{9}{2}$ = $4.5$.