Answer
The geometric series is divergent
Work Step by Step
$\Sigma^{\infty}_{n=1}ar^{n-1} = a+ar+ar^{2}$
If $|r| \lt 1$ the geometric series is convergent
$\Sigma^{\infty}_{n=1}ar^{n-1} = \frac{a}{1-r}$
$|r| \lt 1$
If $|r| \geq 1$ the geometric series is divergent
$\Sigma^{\infty}_{n=1} \frac{10^{n}}{(-9)^{n-1}}$
rewrite $n$-th term of the geometric series in the form $ar^{n-1}$
$=\Sigma^{\infty}_{n=1}10\frac{10^{n-1}}{(-9)^{n-1}}$
$=\Sigma^{\infty}_{n=1}10(\frac{10}{-9})^{n-1}$
$=\Sigma^{\infty}_{n=1}10(-\frac{10}{9})^{n-1}$
First term, $a=10$ and the common ratio is $r= -\frac{10}{9}$
$|r|=|-\frac{10}{9}|$
$=\frac{10}{9}$
$=1.1111$
$\gt 1$
Since $|r| \gt 1$, the geometric series is divergent.