Answer
The infinite geometric series converges.
$S_{\infty} = \dfrac{8}{3}$
Work Step by Step
RECALL:
(1)
In the infinite geometric series:
$$\sum_{k=1}^{\infty}c \cdot r^{n-1}$$
$r$ is the common ratio.
(2)
A geometric series converges if $|r| \lt 1$. The sum of a convergent infinite geometric series is given by the formula:
$S_{\infty}=\dfrac{a_1}{1-r}$
where
$r$ = common ratio
$a_1$ = first term
$\bf\text{Solve for r}:$
Note that when a geometric series is summation notation, the expression being raised to a power is the common ratio.
Thus, the common ratio of the given series is $-\dfrac{1}{2}$.
Since $|-\frac{1}{2}|=\frac{1}{2} \lt 1$, the series converges.
$\bf\text{Find the sum:}$
The first term of the series can be evaluated by substituting $1$ for $k$:
$a_1 = 4\left(-\dfrac{1}{2}\right)^{1-1} = 4\left(-\dfrac{1}{2}\right)^0=4(1) = 4$
With $a_1=4$ and $r=-\dfrac{1}{2}$, solve for the sum using the formula in part (2) above to obtain:
$S_{\infty} = \dfrac{a_1}{a-r}
\\S_{\infty}=\dfrac{4}{1-(-\frac{1}{2})}
\\S_{\infty}=\dfrac{4}{\frac{2}{2}+\frac{1}{2}}
\\S_{\infty}=\dfrac{4}{\frac{3}{2}}
\\S_{\infty}=4 \cdot \dfrac{2}{3}
\\S_{\infty}=\dfrac{8}{3}$
