Simplifying numerater/denomenator exponents for Geometric Series

Question

Take the following series:

$$\sum_{n=1} \frac{(-3)^{n-1}}{4^n}$$

It's pretty easy to find the common ratio to be $-\frac{3}{4}$ by calculating the first few terms and diving them each by the preceding term (and my text book bears this out). However while that value makes intuitive sense given the original formula I can't seem to find a way to justify that algebraically.

I assume it must have something to do with being able to write

$$\frac{(-3)^{n-1}}{4^n}$$

in the form

$$(\frac{-3}{4})^x$$

where $x$ is some simplification of the original exponents, but I can't find a way to "extract" them without breaking any rules and while still getting the correct series values.

Answer 1

We can write the terms of the series as

$$\frac{(-3)^{n-1}}{4^n}=-\frac13 \frac{(-3)^n}{4^n}=-\frac13\left(\frac{-3}{4}\right)^n$$

Thus, we have

$$\sum_{n=1}^{\infty}\frac{(-3)^{n-1}}{4^n}=-\frac13\sum_{n=1}^{\infty}\left(\frac{-3}{4}\right)^n=-\frac13\frac{(-3/4)}{1-(-3/4)}=\frac17$$

Answer 2

As an alternative to Dr. MV's very nice answer, note that the series can be written as $$ \begin{align} \sum_{n=1}^\infty\frac{(-3)^{n-1}}{4^n} &=\frac14\sum_{n=1}^\infty\left(-\frac34\right)^{n-1}\\ &=\frac14\sum_{n=0}^\infty\left(-\frac34\right)^n\\ &=\frac14\frac1{1-\left(-\frac34\right)}\\ &=\frac17 \end{align} $$ This is the way I originally approached this.