Investigate for convergence in $z\in\mathbb{C}, \ |z|<1$: $\Sigma_{k=1}^{+\infty}\left(\cos (z^k)-1\right).$ 
Investigate for convergence in $z\in\mathbb{C}, \ |z|<1$:
$$\Sigma_{k=1}^{+\infty}\left(\cos (z^k)-1\right).$$

We were recommended to use $\cos(z^k)$ expansion first:
$$
\Sigma_{k=1}^{+\infty}\left(\cos (z^k)-1\right)=\Sigma_{k=1}^{+\infty}\Sigma_{n=1}^{+\infty}\left(\frac{(-1)^n z^{2kn}}{(2n)!}\right) =
$$
here we need to reference some theorem that says that $n$ and $k$ changing places $\Sigma_{k=1}^{+\infty}\Sigma_{n=1}^{+\infty}\left(\frac{(-1)^n z^{2kn}}{(2n)!}\right)=\Sigma_{n=1}^{+\infty}\Sigma_{k=1}^{+\infty}\left(\frac{(-1)^n z^{2kn}}{(2n)!}\right)$ is allowed. I don't remember such theorem, and failed to google it. I don't see why $n$ and $k$ changing places would not always be okay. I would be very grateful if someone could give a statement of the theorem.
$$=\Sigma_{n=1}^{+\infty}\frac{(-1)^n}{(2n)!}\Sigma_{k=1}^{+\infty} 
 \ z^{2kn}=\Sigma_{n=1}^{+\infty}\left(\frac{(-1)^n}{(2n)!}\cdot\frac{z^{2n}}{1-z^{2n}}\right)$$
Ratio test:
$$\frac{(-1)^{n+1} z^{2n+2}}{(2n+2)! \ (1-z^{2n+2})} \cdot \frac{(2n)! \ (1-z^{2n})}{(-1)^n \ z^{2n}} 
= 
\frac{(-1) z^{2}(1-z^{2n})}{(2n+1)(2n+2) \ (1-z^{2n+2})} \longrightarrow_{n \longrightarrow \infty} 0 < 1, $$ so the original series is convergent in the given region.
I feel like my solution is wrong. If it is, could someone please point out mistakes, and give hints or explain how to solve correctly?
Thank you.
 A: What's wrong with using the ratio test as-is? As long as $z\neq0$, in which case convergence is trivial, we get:
$$\lim_{k\to\infty}\left|\frac{\cos(z^{k+1})-1}{\cos(z^k)-1}\right|=\lim_{k\to\infty}\left|\frac{-\frac{1}{2}z^{2k+2}+o(z^{4k+3})}{-\frac{1}{2}z^{2k}+o(z^{4k-1})}\right|=\left|\lim_{k\to\infty}\frac{z^2+o(z^{2k+3})}{1+o(z^{2k-1})}\right|=0<1$$
As $z^m\to0,\,m\to\infty$.
As for the interchange, I suspect it is only valid when you already know the series to be absolutely convergent, else you could run afoul of the Riemann series pathologies. Interchanges are generally not valid when using them to ascertain convergence - only after convergence has already been ascertained!
A: Note that we have for $|z|<1$
$$\begin{align}
\left|\sum_{k=1}^K (\cos(z^k)-1)\right|&\le \sum_{k=1}^K |1-\cos(z^k)|\\\\
&=\sum_{k=1}^K \left|\sum_{n=1}^\infty \frac{(-1)^{n}z^{2kn}}{(2n)!}\right|\\\\
&\le \sum_{k=1}^K \sum_{n=1}^\infty \frac{|z|^{2kn}}{(2n)!}\\\\ 
&=\sum_{n=1}^\infty \frac{1}{(2n)!}\frac{|z|^{2n}(1-|z|^{2Kn})}{1-|z|^{2n}}\\\\
&\le \frac{1}{1-|z|^2}\sum_{n=1}^\infty \frac1{(2n)!}\\\\
&<\infty
\end{align}$$
and the series converges as was to be determined.
A: Since $\cos z$ is analytic at $z=0$, there is $M>0$ such that
$$\Big|\frac{\cos(z)-1}{z}\Big|\leq M$$
for all $|z|\leq 1$.
Notice that
$$F(z):=\sum_n(\cos(z^n)-1)=\sum_n\frac{\cos z^n-1}{z^n} z^n$$
Then, by comparison, converges absolutely and uniformly  for all $z$ in a compact subset of the unit disc $|z|<1$: $|\cos z^n -1|\leq M|z|^n$.
