Uniform convergence of $\sum^n_{k=-n} \frac{1}{z+k}$

Question

Let $D=\mathbb C \setminus \mathbb Z$ and define $$f_n(z)=\sum^n_{k=-n}\frac{1}{z+k}$$ I have to prove that $\{f_n\}^\infty_{n=0}$ is locally convergent on D. We are given the hint to write $f_n$ as a nth partial sum of a series en show that that series is normal convergent. I don't really understand this hint however. Wouldn't you get a double summation if I were to try this and if so how would normal convergence of that series imply normal convergence for the sequence?

Answer 1

You can show that $$\sum_{k=-n}^n\frac{1}{z+k}=\frac1z+2z\sum_{k=0}^n\frac1{z^2-k^2}.$$ Now for fixed $z$ whenever $k>\sqrt2|z|$ the reverse trangle inequality you can find that $$\left|\frac1{z^2-k^2}\right|<\frac2{k^2}$$ which is absolutly convergent (and thus so are all it's tails). This shows that the series is normally convergent which implies local convergence (by Weierstrass majorant test).