Series convergence $\frac{(z+2i)^n+(z-i)^n}{n^2 \cdot 2^n}$ Could you tell me what can I do to determine convergence of $\frac{(z+2i)^n+(z-i)^n}{n^2 \cdot 2^n}$ for $z \in \mathbb{C}$ ?
Thank you.
 A: The first thing you should try when you see a sum is to prove the convergence of each term separately:
$\cfrac{(z+2i)^n}{n^2 \cdot 2^n}=\cfrac{1}{n^2}\left(\cfrac{z+2i}{2}\right)^n$
which tells you
$\cfrac{(z+2i)^n}{n^2 \cdot 2^n} \text{ converges} \Leftrightarrow \left|\cfrac{z+2i}{2}\right|\le 1 \Leftrightarrow |z-(-2i)| \le 2$

The same thing for the other operand tells you that
$\cfrac{(z-i)^n}{n^2 \cdot 2^n} \text{ converges} \Leftrightarrow \left|\cfrac{z-i}{2}\right|\le 1 \Leftrightarrow |z-i| \le 2$

If $z$ is in both disks, then you sequence converges. If it is in one of the disks, but not in the other disk, then it diverges. If it is in neither of the disks, we don't know yet.

(From this point on, it's just me rewriting what Did said in the comment)
If both diverge (we are outside of both disks), the only way the sum could converge would be to have $|z+2i|=|z-i|>2$ which means $\exists x \in \Bbb R, z=x-\cfrac{1}{2}i$ with $|x|>\cfrac{\sqrt{7}}{2}$
$\cfrac{(z+2i)^n+(z-i)^n}{n^2 \cdot 2^n}=\cfrac{\left(x+\cfrac{3}{2}i\right)^n+\left(x-\cfrac{3}{2}i\right)^n}{n^2 \cdot 2^n}=\cfrac{2\Re\left(\left(x+\cfrac{3}{2}i\right)^n\right)}{n^2 \cdot 2^n}=\Re\left(\cfrac{2}{n^2}\left(\cfrac{x+\cfrac{3}{2}i}{2}\right)^n\right)=\Re\left(\cfrac{2}{n^2}\left(\cfrac{x}{2}+\cfrac{3}{4}i\right)^n\right)$
But since $\left|x+\cfrac{3}{2}i\right| > 2$, $\cfrac{2}{n^2}\left(\cfrac{x+\cfrac{3}{2}i}{2}\right)^n$ diverges so its real part or its imaginary part or both diverge.

Suppose the real part converges.
Write $x+\cfrac{3}{2}i=re^{i\theta}$
We get that $r^n\cos(n\theta)$ converges where $r>2$ so $\cos(n\theta)\to 0$
We know that $\cos(n\theta + \theta)=\cos(n\theta)\cos(\theta)-\sin(n\theta)\sin(\theta)$
By taking the limit as $n\to +\infty$, we get
$\sin(n\theta)\sin(\theta)\to 0$ Since $\Im(re^{i\theta})=\cfrac{3}{2}\not = 0$, we know that $\sin(\theta)\not=0$
So we get that $\sin(n\theta)\to 0$ so $e^{in\theta}=\cos(n\theta)+i\sin(n\theta)\to 0$ which is absurd since $\left|e^{in\theta}\right|=1$
So the real part diverges.

To conclude, it $z$ is in both disks, the sequences converges and if $z$ is outside at least one of the disks, it diverges.
