Let $U := \{z \in \mathbb{C}: $ Re$(z) > 1\}$ and $f_n: U \rightarrow \mathbb{C}, z \mapsto \frac{(-1)^n}{z + n}$. I'd like to show that $\sum_{k\geq0} g_k$ is (locally) normally convergent on $U$, where $g_k = f_{2k} + f_{2k + 1}$. My first idea has been to find a lower bound for the denominator of $g_k$ as the following:
$\sum_{k\geq0} ||{f_{2k} + f_{2k + 1}}|| = \sum_{k\geq0} ||\frac{1}{z^2 + (4k + 1)z + 4k^2 + 2k}|| = \sum_{k\geq0} \frac{1}{\inf |z^2 + (4k + 1)z + 4k^2 + 2k|} \leq \sum_{k\geq0} \frac{1}{\inf (\text{ Re}(z)^2 - \text{ Im}(z)^2 + (4k +1) \text{ Re}(z) + 4k^2 + 2k)} \leq \sum_{k\geq0} \frac{1}{\inf (- \text{ Im}(z)^2 + 4k^2 + 2k)} \leq \text{ ??} \leq \sum_{k\geq0} \frac{1}{k^2}$
where I used the definition of $U$ and the inequality $|z| \geq \text{ Re}(z)$, which holds for every complex number $z$. My problem: I don't see how to find a lower bound for the term $-\text{ Im}(z)^2$.
Does anyone see how to find such a lower bound?
(Even if I try to consider $g_k (z)$ only on a disk around a point $z_0 \in U$, I have a term $-\text{ Im}(z_0)^2$ and don't see how to conclude)
You have $$ |g_k(z)| = |f_{2k}(z) + f_{2k + 1}(z)| = \left|\frac{1}{z+2k} - \frac{1}{z+2k+1}\right| = \frac{1}{|z+2k| \cdot|z+2k+1|} \, . $$ Now use $$ |z + 2k| \ge \operatorname{Re}(z+2k) = \operatorname{Re}(z) + 2k > 1+2k $$ to conclude that $$ |g_k(z)| \le \frac{1}{(1+2k)(2+2k)} \le \frac{1}{2k^2} \, . $$
(You were actually very close. The difference to your approach is that the factors $(z+2k)$ and $(z+2k+1)$ are estimated seperately, instead of trying to estimate their product.)