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)

$$\begin{align}
\sum_{n=1}^{\infty } \frac{-3^{n-1}}{7^{n}}
& = - \frac{1}{7} \sum_{n=1}^{\infty } (\frac{3}{7})^{n-1}
\\
& = - \frac{1}{7} \frac{1}{1-\frac 3 7}
\\
& = - \frac 1 4
\\[2ex]
\sum_{n=1}^{\infty } \frac{(-3)^{n-1}}{7^{n}}
& = \frac{1}{7} \sum_{n=1}^{\infty } (-\frac{3}{7})^{n-1}
\\
& = \frac{1}{7} \frac{1}{1+\frac 3 7}
\\
& = \frac 1 {10}
\end{align}$$
