Prove that each of the following series converges uniformly on the corresponding subset of $\mathbb C$:
$$\begin{align*} \text{(a)} \; & \sum_{n=1}^\infty \frac{1}{n^2 z^{2n}}, & & \text{on } \{ z: |z| \geq 1\} \\ \text{(b)} \; & \sum_{n=1}^\infty \sqrt{n}\,e^{-nz}, & & \text{on } \{ z: 0 < r < \operatorname{Re}(z)\} \\ \text{(c)} \; & \sum_{n=1}^\infty \frac{2^n}{z^n + z^{-n}}, & & \text{on } \{ z: |z| \leq r < 1/2\} \\ \text{(d)} \; & \sum_{n=1}^\infty 2^{-n} \cos nz, & & \text{on } \{ z: \left|\operatorname{Im}(z)\right| \leq r < \log 2\} \end{align*}$$ Weierstrass M-test: compare with the series (a) $\sum n^{-2}$, (b) $\sum \sqrt{n}\,e^{-nr}$, (c) $\sum (2r)^n/(1-r^{2n})$, (d) $\sum 2^{-n} e^{nr}$.
I am struggling to obtain inequalities for (c) and (d) (the inequalities that I need to get to are listed below) Could someone please help me?
Also if someone could suggest a good website to revise inequalities of this particular level, that would be great!
For (c)
$$\left|\frac{2^n}{z^n + z^{-n}}\right|=\left|\frac{2^nz^n}{z^{2n} + 1 }\right|< \frac{2^n|z|^n}{1 -|z^{2n}|}<\frac{2^nr^n}{1 -r^{2n}}$$
since $|z| \leq r < 1/2$ and $0 < 1 - r^{2n} < |1|-|z^{2n}|\leq |1- (-z^{2n})|=|1+z^{2n}|$