I am given the following geometric series and am asked to find the sum.
$$\sum_{n=1}^{\infty} \left(\frac{12}{(-5)^n}\right)$$
I know that I somehow need to get this in the form $\sum_{n=1}^{\infty}ar^{n-1}$, where $a$ is the first term and $r$ is the ratio, but the best I could come up with is the following:
$$\sum_{n=1}^{\infty} \left(12(-5)^{-n}\right)$$
However, It needs to be in the following form:
$$\sum_{n=1}^{\infty} ar^{n-1} = \frac{a}{1-r}$$
I am not sure what I am missing here… can someone point me in the right direction?
Best Answer
$$12\;(-5)^{-n}=12\;\left(-\frac{1}{5}\right)^n=-\frac{12}{5}\left(-\frac{1}{5}\right)^{n-1}$$