<s>[INST] 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 $r$ = -$1 \over 4$
 [/INST] 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 $S_{odd}$:
$$ \begin{align*}S&= \frac{16}{15} - \frac{4}{15}\\ S&= \frac{12}{15}\\ S&=\frac{4}{5}\end{align*}$$
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