Answer
series converges;
$S_{\infty} = 12$
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 $\dfrac{1}{3}$.
Since $|\frac{1}{3}|=\frac{1}{3} \lt 1$, the series converges.
$\bf\text{Find the sum:}$
The first term of the series can be evaluated by substituting $1$ for $k$:
$a_1 = 8\left(\dfrac{1}{3}\right)^{1-1} = 8\left(\dfrac{1}{3}\right)^0=8(1) = 8$
With $a_1=8$ and $r=\dfrac{1}{3}$, solve for the sum using the formula in part (2) above to obtain:
$S_{\infty} = \dfrac{a_1}{a-r}
\\S_{\infty}=\dfrac{8}{1-\frac{1}{3}}
\\S_{\infty}=\dfrac{8}{\frac{3}{3}-\frac{1}{3}}
\\S_{\infty}=\dfrac{8}{\frac{2}{3}}
\\S_{\infty}=8 \cdot \dfrac{3}{2}
\\S_{\infty}=\dfrac{24}{2}
\\S_{\infty}=12
$
