Compact convergence of series

Question

I am trying to show that

$$f(z) = \sum_{k=1}^{\infty} \frac{(-1)^{k}}{z+k}$$

converges compactly over $\mathbb{C}$ and starting to think that this statement is false after several attempts.

If I take an $\epsilon >0$ and try to an $N$ so that for $n\geq N$,

$$|| \sum_{k=1}^{n} \frac{(-1)^{k}}{z+k} - f(z)|| < \epsilon,$$

I get stuck. The $z+k$ is preventing me from putting a bound on the function in a way that is useful. I've tried using the Weierstrass M-test, but I couldn't make it work either.

Answer 1

Hint: You can write the series as $$ \sum_{k=1}^\infty\frac1{z+2k}-\frac1{z+2k-1}=-\sum_{k=1}^\infty\frac1{(z+2k)(z+2k-1)} $$ which does converge absolutely.