Answer
convergent;
$\S_{\infty}=\dfrac{1}{648}$
Work Step by Step
RECALL:
(1) The sum of an infinite geometric series is convergent if $|r| < 1$. The sum is given by the formula:
$S_{\infty}=\dfrac{a}{1-r}$
(2) An infinite geometric series is divergent if $|r|>1$.
(3) The common ratio $r$ of a geometric series can be found by dividing any term by the term before it:
$r = \dfrac{a_n}{a_{n-1}}$
Solve for $r$ to obtain:
$\require{cancel}
r = \dfrac{\frac{1}{3^8}}{\frac{1}{3^6}}
\\r=\dfrac{1}{3^8} \cdot \dfrac{3^6}{1}
\\r=\dfrac{3^6}{3^8}
\\r=\dfrac{\cancel{3^6}}{\cancel{3^8}3^2}
\\r=\dfrac{1}{3^2}=\dfrac{1}{9}$
Since $|\frac{1}{9}|<1$, then the series converges.
Solve for the sum using the formula above, with $a=\dfrac{1}{3^6}$ and $r=\frac{1}{9}$, to obtain:
$\require{cancel}
S_{\infty}=\dfrac{\frac{1}{3^6}}{1-\frac{1}{9}}
\\S_{\infty}=\dfrac{\frac{1}{3^6}}{\frac{9}{9}-\frac{1}{9}}
\\S_{\infty}=\dfrac{\frac{1}{3^6}}{\frac{8}{9}}
\\S_{\infty}=\dfrac{\frac{1}{3^6}}{\frac{8}{3^2}}
\\S_{\infty} = \dfrac{1}{3^6} \cdot \frac{3^2}{8}
\\S_{\infty} = \dfrac{1}{\cancel{3^6}3^4} \cdot \dfrac{\cancel{3^2}}{8}
\\S_{\infty}=\dfrac{1}{648}$
