Determine the locally uniform convergence of the series $\sum_{n=1}^{\infty} f_n(z)$ on $V=\mathbb{C}\setminus\{-n : n\in\mathbb{N}\}$, where $$f_{n}(z) = (-1)^n \frac{1}{z+n}.$$ Is $f_n$ also uniform convergent?
My idea: Let be $a_n(z):=(-1)^n$ and $g_n(z):=\frac{1}{z+n}$. Then $s_n:=\sum_{n=1}^{\infty} a_n(z)$ is bounded, $(f_n)_{n\in\mathbb{N}}$ converges uniformly to the null function. I need to show that $\sum_{n=1}^{\infty}|f_{n+1}-f_{n}|=\sum_{n=1}^{\infty}|\frac{1}{(z+n+1)(z+n)}|$ converges uniformaly to conclude that the series $\sum_{n=1}^{\infty} f_n(z)$ converges uniformally on V. But how do I now show that $\sum_{n=1}^{\infty}|f_{n+1}-f_{n}|$ is uniformly convergent? It seems that convergence tests don't work, and I don't find it not very easy to compare it. What to do next? Is this the right idea? Thanks in advance!
That series of functions converges uniformly on any compact subset $K$ of $\Bbb C\setminus\{-n\mid n\in\Bbb N\}$. In fact, if $M=\max_{z\in K}|z|$, you have, when $n>M$,$$|(z+n+1)(z+n)|\geqslant\bigl((n+1)-M\bigr)(n-M),$$and therefore$$\left|\frac1{(z+n+1)(z+n)}\right|\leqslant\frac1{\bigl((n+1)-M\bigr)(n-M)}.\tag1$$So, $\sum_{n=1}^\infty\frac1{((n+1)-M)(n-M)}$, it follows from $(1)$ and the Weierstrass $M$-test that your series converges uniformly on $K$.