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 $\frac{1}{7}$

If $\,a, ar, ar^2,...\,$ is a geometric series with $\,|r|<1\,$ ,then
$$\sum_{n=0}^\infty ar^n=\lim_{n\to\infty} ar^n=\lim_{n=0}\frac{a(1-r^n)}{1-r}=\frac{a}{1-r}$$since $\,r^n\xrightarrow [n\to\infty]{} 0\Longleftrightarrow |r|<1\,$ , and thus
$$\sum_{n=1}^\infty\frac{(-3)^{n-1}}{4^n}=\frac{1}{4}\sum_{n=0}^\infty \left(-\frac{3}{4}\right)^n=\frac{1}{4}\frac{1}{1-\left(-\left(\frac{3}{4}\right)\right)}=\frac{1}{4}\frac{1}{\frac{7}{4}}=\frac{1}{7}$$
