I am doing some self-studying (and preparing for an exam) and I'm confronted with this task:
$$S_N(x)=\sum_{k=-N}^N\frac{1}{x-k}, x \in \mathbb{R} \setminus \mathbb{Z}$$
How can I show that $S_N(x)$ converges absolutely on every compact subset $[a,b] \subset \mathbb{R} \setminus \mathbb{Z}$ against a function $h(x)$?
The hint I received is that $S_N(x)=\frac{1}{x}+2x \sum_{k=1}^N\frac{1}{x^2-k^2}$ (I don't know how I can prove that though). I could already show that
$\sum_{k=K+1}^\infty \frac{1}{(x-k)^2}$ and $\sum_{k=K+1}^\infty \frac{1}{(x+k)^2}$ are absolutely convergent for $x \in \mathbb{R} \setminus \mathbb{Z}, x\in[-K,K]$(Weierstraß M-test) from which follows that
$f(x)=\sum_{k=K+1}^\infty \frac{1}{(x-k)^2}+\sum_{k=K+1}^\infty \frac{1}{(x+k)^2}+\sum_{k=-K}^K \frac{1}{(x-k)^2}$ ist continuous
but I don't know if that helps
Fix $x>0$ such that $x\in[a,b]\subset(n,n+1)$ for some $n\in\mathbb{N}_0$. Then, for $N>n+1$, $$ S_N(x)=\frac{1}{x}+2x \sum_{k=1}^N\frac{1}{x^2-k^2}=\frac{1}{x}+2x \sum_{k=1}^{n}\frac{1}{x^2-k^2}+2x \sum^{N}_{k=n+1}\frac{1}{x^2-k^2}. $$ Since $$ \left|\sum^{N}_{k=n+1}\frac{1}{x^2-k^2}\right|=\sum^{N}_{k=n+1}\frac{1}{k^2-x^2}\le\sum^{N}_{k=n+1}\frac{1}{k^2-b^2}$$ and $\sum^{N}_{k=n+1}\frac{1}{k^2-b^2}$ converges as $N\to\infty$, one has that $\sum^{N}_{k=n+1}\frac{1}{x^2-k^2}$ converges uniformly in $[a,b]$. Hence $S_N(x)$ converges uniformly to some function $h(x)$ in $[a,b]$. Since $S_N(x)$ is continuous in $[a,b]$, so is $h(x)$.