Convergence of $\sum_{n =1}^\infty\frac{\cos(iz)^n}{n}$

I want to check convergence of
$$\sum_{n =1}^\infty\frac{ \cos(iz)^n}{n}$$

My work so far

Let $x = \cos(iz)$. Then we are interested in series $\sum_{n=1}^\infty\frac{x^n}{n}$

Assume that $a_n = \frac 1 n$. We have that reverse of convergence radius equals to:

$$\frac 1 R = \lim_{n \rightarrow \infty} \frac{|a_{n+1}|}{|a_n|} = \lim \frac{n}{n+1} = 1 \Rightarrow R = 1$$

We see then that it will converge for $x \in B(0, 1)$ i.e. $|x| < 1 \Leftrightarrow|\cos(iz)| < 1 \Leftrightarrow |\cos(ix - y)| < 1$

We know that

$$|\cos(x + iy)|^2 = \cos^2x + \sinh^2y$$

So we have that series converges when:

$$\sqrt{\cos^2(-y) + \sinh^2x} < 1 \;\;\;\;\;\;\;\;\;\;\; (1)$$

And diverge when $\sqrt{\cos^2(-y) + \sinh^2x} > 1$. We have to manually check case when radius equals to $1$. In other words $|x| = 1 \Rightarrow x = e^{i\alpha}$

We are interested in sum $$\sum_{n =1}^\infty \frac{e^{ni\alpha}}{n}$$

I wanted to show somehow that this series converge. I was thinking about Dirichlet test - the only thing that is not obvious is boundness of partial sums: $\sum_{n =1}^N e^{ni\alpha}$.

In other words I want to show that:

$$\exists_{M \in R}: \sum_{n = 1}^N e^{n i \alpha} \le M \Leftrightarrow e^{i \alpha} \cdot \frac{1 - e^{N i \alpha}}{1 - e^{i \alpha}} \le M \;\;\;\;\;\;\;\; (2)$$

But I'm not sure how can I prove it.

Questions

$(1)$ I have question about my condition of convergence i.e. $\sqrt{\cos^2(-y) + \sinh^2x} < 1$. Can I simplify it somehow?

$(2)$ Is this sequence bounded? If yes, could you please help me proving so?

Solution 1:

If $e^{i\alpha}=1$ then $ \sum\limits_{k=1}^{n} \frac { e^{ik\alpha}} n=\sum\limits_{k=1}^{n} \frac 1 n$ which is unbounded. Also, $ \sum\limits_{k=1}^{\infty} \frac { e^{ik\alpha}} n$ diverges . If $e^{i\alpha}\neq 1$ then $|e^{i \alpha} \frac {1-e^{iN\alpha}} {1-e^{i\alpha} }| \leq \frac {1+1} {|1-e^{i\alpha}}|$.

For (1) I don't think you can simplify it further.