The question is prove that the series $\sum_{n=0}^{\infty} \frac{(-1)^n}{z+n}$ determines a meromorphic function. So the way that I prove these kinds of questions is fix $R>0$ and prove that it is meromorphic at any disk $D(0,R)$. Then let $N>2R$ and $g(z)=\sum_{n=0}^{N} \frac{(-1)^n}{z+n}$ and $h(z)=\sum_{n=N+1}^{\infty} \frac{(-1)^n}{z+n}$. Then g is clearly meromorphic and I need to prove that h is holomorphic and then I am done. To that end I need to prove that the sum in $h$ converges uniformly. In the previous exercises I have tried so far, I could use M-test to determine that, which here clearly doesn't work, as it's not absolutely convergent.
So I am curious on how to check uniform convergence of alternating series in complex analysis? Is there a useful test for that? Or do I have to just get my hands dirty with just proving that it is Cauchy?
I was wondering whether the following 2 statements are correct:
(i) If $\sum_{n=0}^{\infty} \frac{(-1)^n}{|z+n|}$ converges uniformly then $\sum_{n=0}^{\infty} \frac{(-1)^n}{z+n}$ converges uniformly.
(ii) If f=$\sum_{n=0}^{\infty} (-1)^nf_n(x)$ a series in the real numbers and $f_n>0$ is decreasing to $0$ uniformly, then $f$ converges uniformly.