Determine whether $\sum\limits_{i=1}^\infty \frac{(-3)^{n-1}}{4^n}$ is convergent or divergent. If convergent, find the sum. $$\sum\limits_{i=1}^\infty \frac{(-3)^{n-1}}{4^n}$$
It's geometric, since the common ratio $r$ appears to be $\frac{-3}{4}$, but this is where I get stuck. I think I need to do this: let $f(x) = \frac{(-3)^{x-1}}{4^x}$.
$$\lim\limits_{x \to \infty}\frac{(-3)^{x-1}}{4^x}$$
Is this how I handle this exercise? I still cannot seem to get the answer $\frac{1}{7}$

Let $q = \frac{-3}{4}$, $a_n = \frac{(-3)^{n-1}}{4^n}$, $b_0 = 0$, $b_n = b_{n-1} + a_n$.
$a_n = -\frac{1}{3} (\frac{-3}{4})^n = -\frac{1}{3} q^n$, hence $b_n = -\frac{1}{3} c_n$, where $c_0 = 0$, $c_n = c_{n-1} + q^n$.
The $c_n$ limit is equal to $q + q^2 + q^3 + \ldots = \frac{q}{1-q} = \frac{\frac{-3}{4}}{1-\frac{-3}{4}} = \frac{-3}{7}$, thus $b_n$ limit is equal to $\frac{1}{7}$.
