Answer
The sum of the series is $S= \frac{1}{7}$
Work Step by Step
$\Sigma^{\infty}_{n=1} \frac{(-3)^{n-1}}{4^{n}}$
$=\Sigma^{\infty}_{n=1} \frac{(-3)^{n-1}}{4 \times 4^{n-1}}$
$=\Sigma^{\infty}_{n=1} (\frac{1}{4}) (-\frac{3}{4})^{n-1}$
First term $a=\frac{1}{4}$ and the common ratio is $r= -\frac{3}{4}$
since $r= -\frac{3}{4} \lt 1$ the series convergent
$S=\frac{a}{1-r}$
$S=\frac{\frac{1}{4}}{1-(-\frac{3}{4})}$
$S=\frac{\frac{1}{4}}{1+\frac{3}{4}}$
$S=\frac{\frac{1}{4}}{\frac{7}{4}}$
$S= \frac{1}{7}$
