Test convergence of the series and find its sum: $\sum_{n=0}^{\infty}\left(\frac{-1-2i}{2+3i}\right)^n$ Test convergence of the series and find its sum: $$\sum_{n=0}^{\infty}\left(\frac{-1-2i}{2+3i}\right)^n$$
My try:
This is simplified to $$\sum_{n=0}^{\infty}\left(\frac{-i-8}{13}\right)^n$$ 
Then how to proceed?

$\sum_{n=0}^{\infty}\left(\frac{-1-2i}{2+3i}\right)^n
$
Since
$\frac{-1-2i}{2+3i}
=\frac{-1-2i}{2+3i}\frac{2-3i}{2-3i}
=\frac{-8-i}{13}
$
and
$|\frac{-8-i}{13}|^2
=\frac{65}{169}
\lt 1
$,
the sum converges.
Applying the standard formula,
the sum is
$\begin{array}\\
\dfrac{1}{1-\frac{-8-i}{13}}
&=\dfrac{13}{13+(8+i)}\\
&=\dfrac{13}{21+i}\\
&=\dfrac{13}{21+i}\dfrac{21-i}{21-i}\\
&=\dfrac{13(21-i)}{442}\\
&=\dfrac{21-i}{34}\\
\end{array}
$
