Show that the series is not absolutely convergent but is uniformly convergent in the whole complex plane

Question

Show that the series is not absolutely convergent but is uniformly convergent in the whole complex plane.

$$\frac{1}{1+|z|}-\frac{1}{2+|z|}+\frac{1}{3+|z|}-\frac{1}{4+|z|}+...+\frac{(-1)^{n-1}}{n+|z|}+...$$

My initial thought was to use the Weierstrass M-test but that does not lead me in the right direction or I do not believe it did.

Answer 1

Update. Since $$|a_n| = \frac {1} {n + |z|} \sim \frac {1} {n},$$ we have $$\sum |a_n| \sim \sum \frac {1} {n},$$ so the series is not absolutely convergent. Also, we have $$|a_n| = \frac {1} {n + |z|} > \frac {1} {n + 1 + |z|} = |a_{n + 1}|.$$ Hence, the series converges by the Leibniz test. But the convergence of the series does not depend on $z$, so the series converges uniformly.

Answer 2

Not absolutely convergent because $$ \sum_{n=1}^\infty{1\over n+|z|}>\sum_{n=1}^\infty{1\over n+\lceil|z|\rceil}=\sum_{n=\lceil|z|\rceil+1}^\infty{1\over n}=\infty $$ Uniformly convergent by alternating series test.