Proving complex series is convergent Question 1: Test the following series for convergence
$$\sum_{n=0}^\infty \left(\frac{2+i}{3-4i}\right)^{2n}$$
I got that it was convergent by doing a ratio test and getting 
$$p = \frac{\sqrt{3}}{5} < 1 $$
which checks out for convergence. Wondering if that was the correct way to approach solving this series.
Question 2:    Same as Question 1
$$\sum_{n=0}^\infty \left(\frac{1+i}{2-i}\right)^{n}$$
I am unsure how to approach solving this question and any advice would be much appreciated. 

The first series is a geometric series of the form $\sum_{n=0}^{\infty} q^n$ with $q = \left( \frac{2 + i}{3 - 4i} \right)^2$. Since
$$ \left| \frac{2 + i}{3 - 4i} \right| = \left| \frac{2 + i}{3 - 4i} \frac{3 + 4i}{3 + 4i}\right| = \left| \frac{2 + 11i}{25} \right| = \frac{\sqrt{4 + 121}}{25} = \frac{1}{\sqrt{5}} $$
we have $|q| = \frac{1}{5} < 1$ and so the series converges.
Similarly, for the second series we have $q = \frac{1 + i}{2 - i}$ and
$$ |q| = \left| \frac{1 + i}{2 - i} \right| = \left| \frac{1 + i}{2 - i} \frac{2+i}{2+i} \right| = \left| \frac{1 + 3i}{5}\right| = \frac{\sqrt{1 + 9}}{5} = \sqrt{\frac{2}{5}} $$
and agan, the series converges.
