Does $\sum_{n\geq 1}\left(\frac{1+2i}{\sqrt{5}}\right)^n$ converge? I am having trouble understanding how to apply convergence tests to complex series. Which test(s) could I use here to determine whether this series converges or diverges? Thanks!
 A: It's a geometric series $\sum_n r^n$ with $r = (1+2i)/\sqrt{5}$. It is guaranteed to converge if $|r|<1$ and diverge if $|r|>1$.
Check if
$$|r| = \left|\frac{1+2i}{\sqrt{5}}\right| < 1.$$

I initially wanted to leave this as a hint, however I'll just go into greater detail. As discovered, you get that $|r|=1$ exactly. Hence the terms
$$\left|\frac{1+2i}{\sqrt{5}}\right|^n = 1 \not\to 0$$
as $n\to\infty$, hence the series diverges by the $n$th term test.
A: $z=\frac{1+2i}{\sqrt{5}}$ is a point on the unit circle, $z=e^{i\arctan 2}$. It follows that
$$ S_N = \sum_{n=1}^{N} z^n = z\cdot\frac{z^N-1}{z-1} $$
has a bounded modulus, but it is not converging as $N\to +\infty$.
A: $$
\frac{1+2i}{\sqrt 5} = \cos\varphi + i \sin \varphi \text{ where } \varphi = \arctan 2.
$$
So
$$
\left( \frac{1+2i}{\sqrt 5} \right)^n = \cos(n\varphi) + i\sin(n\varphi).
$$
The terms remain on the circle $|z|=1$ as $n$ changes. Therefore the terms do not approach $0.$ Therefore the series diverges.
(However it does have a Cesàro sum.)
