Find the domain of analyticity of a function $$\sum_{n=0}^{+\infty}\frac{(-1)^n}{z-n}_, \ \ \ z\in\mathbb{C}.$$
If there's uniform convergence, there's also analyticity. Let's use Weierstrass's test to investigate uniform convergence of the series. Intuitively, it seems to be convergent $\forall z.$
$$ \sum_{n=0}^{+\infty}\left|\frac{(-1)^n}{z-n} \right| = \sum_{k=0}^{+\infty} \left| \frac{1}{z-2k}-\frac{1}{z-2k-1} \right| = \sum_{k=0}^{+\infty}\left|\frac{z-2k-1-z+2k}{(z-2k)(z-2k-1)}\right| = \sum_{k=0}^{+\infty}\left|\frac{1}{(z-2k)(z-2k-1)} \right| = \sum_{k=0}^{+\infty}\left|\frac{1}{(z-2k)^2-z+2k} \right| \leq\sum_{k=0}^{+\infty}\left|\frac{1}{(z-2k)(z-2k-1)} \right| \leq $$
Since some $K\in \mathbb{N},$ $|z-2k-1|>|z-2k|, k>K.$
$$ \leq C + \sum_{k=K+1}^{+\infty} \frac{1}{|z-2k|^2} = C + \sum_{k=K+1}^{+\infty} \frac{1}{|(x-2k)^2+y^2|} \leq C + \sum_{k=K+1}^{+\infty} \frac{1}{|-kx+k^2|}. $$
If $x\leq0,$ then $$C + \sum_{k=K+1}^{+\infty} \frac{1}{|-kx+k^2|} \leq C + \frac{\pi^2}{6},$$ and the original series is analytical on the right half-plane.
If $x > 0$, then for large enough $k > M$
$$ C + \sum_{k=K+1}^{+\infty} \frac{1}{|-kx+k^2|} = C + C_1 + \sum_{k=M+1}^{+\infty} \frac{1}{k^2-kx}. $$
I am stuck here. I am not sure my thoughts are correct, and will be grateful for any help, including alternative solutions. Thank you.
The best way to do this is to write $$\sum_{n=0}^{N}\frac{(-1)^n}{z-n}=\frac{1}{z}+\sum_{n=1}^{N}(\frac{(-1)^n}{z-n}+\frac{(-1)^n}{n})+x_n$$ where $x_n=-\sum_{n=1}^{N}\frac{(-1)^n}{n} \to \log 2$
Now it is clear that the term inside the bracket converges locally uniformly on $\mathbb C-\mathbb N^*$ so defines an analytic function there, while the $1/z$ term is analytic on $\mathbb C^*$ giving the result that the series defines an analytic function except at the nonnegative integers