# How can I find the sum of the series $\sum_{n=0}^\infty {(-1)^n \over 4^n}$ or show that it diverges using the geometric series test?

First, I reindexed it:

$$\sum_{n=0}^\infty {(-1)^n \over 4^n} = \sum_{n=1}^\infty {(-1)^{n-1} \over 4^{n-1}} = \sum_{n=1}^\infty {\left(-1 \over 4\right)}^{n-1}$$

So now I'm pretty sure it's in the form $ar^{n-1}$, but using the formula $\left(a \over {1-r}\right)$, I'm not sure I understand what $a$ would be, I'm guessing that $r$ = -$1 \over 4$

Yes you are right, the ratio is $-\frac{1}{4}$. Since the absolute value of ratio is less then one, then the series converge. Actually the reindexing is not necessary. Rather, without reindexing, the expression is neater. The series converge to $$\frac{1}{1-r}=\frac{1}{1+\frac{1}{4}}=\frac{5}{4}$$ You can check the wikipedia page.

If we break the summations down we get:

$$S = \sum_{i=0}^{n}\frac{1}{4^{2(n-1)}} - \sum_{i=0}^{n}\frac{1}{4^{2n-1}}$$

That is the sum of the reciprocals of 4 to the power of the even positive numbers, then you subtract the reciprocals of 4 to the power of the odd positive numbers. $2(n-1)$ is the $n_{th}$ even number, and likewise $2n-1$ is the $n_{th}$ odd number. If take the summation with the even numbers: $$S_{even} = \frac{1}{4^0} + \frac{1}{4^2} + ... \frac{1}{4^{2(n-1)}}$$ Multiply by the common factor $\frac{1}{4^2}$: $$\frac{S_{even}}{4^2} = \frac{1}{4^2} + \frac{1}{4^4} + ... \frac{1}{4^{2n}}$$ Now subtract the multiplied sum: $$S_{even} - \frac{S_{even}}{4^2} = \frac{1}{4^0} -\frac{1}{4^{2n}}$$ We can see that all of the terms cancel each other out, except for the first term and the last term in the second summation. We can also see that: $$\lim\limits_{n \to \infty} \frac{1}{4^{2n}} = 0$$ So: $$S_{even} - \frac{S_{even}}{4^2} = \frac{1}{4^0} - 0$$ Rearrange: $$S_{even} = \frac{16}{15}$$

Now we do the same for the odd summation: $$S_{odd} = \frac{1}{4^1} + \frac{1}{4^3} + ... \frac{1}{4^{2n-1}}$$ Multiply through by the common factor ,then subtract that from the original: $$S_{odd} - \frac{S_{odd}}{4^2} = \frac{1}{4^1} -\frac{1}{4^{2n+1}}$$ But: $$\lim\limits_{n \to \infty} \frac{1}{4^{2n+1}} = 0$$ So: $$S_{odd} - \frac{S_{odd}}{4^2} = \frac{1}{4^1} - 0$$ Rearrange: $$S_{odd} = \frac{4}{15}$$ Go back to the original series and substitute in $S_{even}$ and $S_{odd}$: \begin{align*}S&= \frac{16}{15} - \frac{4}{15}\\ S&= \frac{12}{15}\\ S&=\frac{4}{5}\end{align*}