Answer
$\dfrac{-3}{5}$
Work Step by Step
The sum of the geometric series is given by $s_n=\dfrac{a}{(1-r)}$
Here, $a$ is initial term and $r$ is common ratio.
A geometric series is to be convergent when $|r|\lt 1$ and divergent when $|r|\gt 1$.
From the given problem, we have $s_n=\Sigma_{n=1}^{\infty}(-1)^n(\dfrac{3}{4^n})$
$r=\dfrac{-1}{4}$ and $a=3(\dfrac{-1}{4})^1=\dfrac{-3}{4}$
Thus,
$s_n=\dfrac{a}{(1-r)}=\dfrac{\dfrac{-3}{4}}{(1-(\dfrac{-1}{4}))}$
Hence, $s_n=\dfrac{-3}{5}$
