Determine the behavior of series on the circles edge

Question

Determine the behavior of series on the circles edge

$A = \sum_{n=1}^{∞} {z^{n!} \over n^2 }$, where $z$ is a complex number

The radius of convergence is $1$ ($l = \lim_{n→∞} \sqrt[n]{1\over {n^2}} = 1 ,R = {1\over {l}} = 1$) , which means $|z| = 1 => z = e^{i\phi }$ and $A = \sum_{n=1}^{∞} {e^{i\phi n!} \over n^2}$

Now I need to show that $\lim_{n→∞} {e^{i\phi n!} \over n^2} = 0$

Here I am stuck. Any help or ideas are welcome!

Answer 1

The absolute value of $ {e^{i\phi n!} \over n^2}$ is $$\vert {e^{i\phi n!} \over n^2} \vert = {\vert e^{i\phi n!} \vert \over n^2} = {1 \over n^2}$$ and $1/n^2$ clearly goes to $0$. Since the absolute value of your sequence goes to $0$, the sequence itself must go to $0$ as well.

Answer 2

In your sequence, the numerator of each term has absolute value $1$, whereas the denominators go to $+\infty$. Therefore, the limit of your sequence is $0$.