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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.

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    $\begingroup$ You could write ${(-3)^{n-1}\over 4^n}$ as $c\cdot\left({-3\over 4}\right)^n$ for a suitable number $c$. $\endgroup$
    – John Dawkins
    Oct 1, 2015 at 22:01
  • $\begingroup$ Could you give an example of a suitable $c$ ? Or in the context of an infinite series would it not matter what $c$ was? $\endgroup$
    – user3776749
    Oct 1, 2015 at 22:18

2 Answers 2

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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

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

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  • $\begingroup$ Yet another time in math where I feel like an absolute cretin :| Stared at that for a few seconds before it clicked. Sincerest thanks. $\endgroup$
    – user3776749
    Oct 1, 2015 at 22:25
  • $\begingroup$ You're welcome. My pleasure. And please, don't be so hard on yourself. All of us make mistakes and we can benefit by learning from them. You'll be fine. $\endgroup$
    – Mark Viola
    Oct 1, 2015 at 22:28
  • $\begingroup$ (+1) This is completely correct, but perhaps I have a bit of OCD since I try to make the series start at $n=0$ so I can use $\sum\limits_{n=0}^\infty r^n=\frac1{1-r}$. $\endgroup$
    – robjohn ♦
    Oct 2, 2015 at 8:53
  • $\begingroup$ @robjon I had thought about showing this solution along with the one you presented. But, I'd leave that for another user. It's always useful pedagalogically to have multiple approaches. Aside, thank you for the up vote! $\endgroup$
    – Mark Viola
    Oct 2, 2015 at 13:32
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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.

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