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 $\frac{1}{7}$

$By the given series have:
$a_{n}=\frac{(-3)^{n-1}}{4^{n}}$,                $a_{n+1}=\frac{(-3)^{n}}{4^{n+1}}$
By the criterion of Dalamber have:
$A=\lim\frac{a_{n+1}}{a_{n}}=\frac{3}{4}<1$
Under this criterion we have that A<1 conclude that given series is convergent.
Since the given series is convergent exist sum of this series. Mark wis $S_{n}$ sum of this series, it is $S_{n}=\sum\limits_{i=1}^\infty \frac{(-3)^{n-1}}{4^n}$.
Hance we
$S_{n}=\sum\limits_{i=1}^\infty \frac{(-3)^{n-1}}{4^n}$
$S_{n}=-\frac{1}{3}\sum\limits_{i=1}^\infty \frac{(-3)^{n}}{4^n}$
$\frac{(-3)^{n}}{4^n}$ is it a geometric series. Now find sum this series. Find the sum must assign a_{1} dhe q.
$a_{1}=-\frac{3}{4}$,    $q=-\frac{3}{4}$.
Sum accounst
$S_{n}=\frac {a_{1}(1-q^{n})}{1-q}=-\frac{3}{7}{[1-\frac{3^{n}}{4^{n}}]}$
$\lim S_{n}=-\frac{3}{7}$
Theres definitely have
$S_{n}=-\frac{1}{3}\sum\limits_{i=1}^\infty\frac{(-3)^{n}}{4^{n}}$
$S_{n}=(-\frac{1}{3})(-\frac{3}{7})$
$S_{n}=\frac{1}{7}$
Conlude the: Given series is the convergent and its sum $\frac{1}{7}$.$
