Summation of Infinite Geometric Series Determine the sum of the following series:
$$\sum_{n=1}^{\infty } \frac{(-3)^{n-1}}{7^{n}} $$
My work:
$$\sum_{n=1}^{\infty } \frac{(-3)^{n-1}}{7^{n}} = \sum_{n=1}^{\infty } \frac{-1}{7} (\frac{3}{7})^{n-1}$$
$$\sum_{n=1}^{\infty } ar^{n-1} = \frac{a}{1-r} = \frac{\frac{-1}{7}}{1-\frac{3}{7}} = -\frac{1}{4}$$
Why does this not work?
Sorry for the incorrect initial post!!!
Edit: -3 changed to (-3)

$$-\frac{1}{3} \sum_{n=1}^{\infty} \frac{3^n}{7^n}=-\frac{1}{3} \sum_{n=1}^{\infty} \left ( \frac{3}{7} \right )^n=-\frac{1}{3} \sum_{n=0}^{\infty} \left ( \frac{3}{7} \right )^n+\frac{1}{3}=-\frac{1}{3} \frac{1}{1-\frac{3}{7}}+\frac{1}{3} \\ =-\frac{1}{3}\frac{7}{7-3}+\frac{1}{3}=-\frac{7}{12}+\frac{4}{12}=\frac{-1}{4}$$
EDIT:
$$-\frac{1}{3} \sum_{n=1}^{\infty} \frac{(-3)^n}{7^n}=-\frac{1}{3} \sum_{n=0}^{\infty} \frac{(-3)^n}{7^n}+\frac{1}{3}=-\frac{1}{3} \frac{1}{1-\left (\frac{-3}{7} \right )}-\frac{1}{3}=-\frac{1}{3} \frac{1}{1+\frac{3}{7}}+\frac{1}{3}=-\frac{1}{3} \frac{7}{10}+\frac{1}{3}=-\frac{7}{30}+\frac{10}{30}=\frac{1}{10}$$
