Answer
convergent;
$S_{\infty}=-\dfrac{1000}{117}$
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 $|r|\ge1$.
(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:
$\require{cancel}
r = \dfrac{\frac{10}{3}}{-\frac{100}{9}}
\\r=\dfrac{10}{3} \cdot \left(-\dfrac{9}{100}\right)
\\r=\dfrac{\cancel{10}}{\cancel{3}} \cdot \left(-\dfrac{\cancel{9}3}{\cancel{100}10}\right)
\\r=-\dfrac{3}{10}$
Since $|-\frac{3}{10}|<1$, then the series is convergent.
Solve for the sum using the formula above, with $a=-\dfrac{100}{9}$ and $r=-\dfrac{3}{10}$, to obtain:
$\require{cancel}
S_{\infty}=\dfrac{-\frac{100}{9}}{1-(-\frac{3}{10})}
\\S_{\infty}=\dfrac{-\frac{100}{9}}{\frac{10}{10}+\frac{3}{10}}
\\S_{\infty}=\dfrac{-\frac{100}{9}}{\frac{13}{10}}
\\S_{\infty} = -\dfrac{100}{9} \cdot \dfrac{10}{13}
\\S_{\infty}=-\dfrac{1000}{117}$
