Prove the series of $$\sum_{n=-\infty}^{\infty}\frac{1}{(z+n)^2}$$ converges uniforply on every set $K \setminus \Bbb{Z}$ such that $K \subseteq \Bbb{C}$ compact.
Here is my attempt:
I proved that the series converges pointwise for every $z \in \Bbb{C} \setminus \Bbb{Z}$
Now since $K$ is compact exists $R>0$ such that $K \subseteq B(0,R)$
We have 2 cases:
$1)$ $K$ contains a finitely many integers,say $m_1,...m_k$.
Note that exists $N \in \Bbb{N}$ such that $N>2R$ so $N>\max\{|m_1|,...,|m_k|\}$ so we have for $z \in K \setminus \Bbb{Z}$ and for $n \geq N$ that $|z+n| \geq n-|z|>n-R>0$ $$\sum_{|n| \geq N}\frac{1}{|z+n|^2} \leq \sum_{|n| \geq N}\frac{1}{(n-|z|)^2}\leq \sum_{|n| \geq N}\frac{1}{(n-R)^2}<\infty,\forall z \in K \setminus \Bbb{Z} $$
By the Weierstrass $M-$test we have that the partial sums $\sum_{N \leq |n| \leq m}\frac{1}{(z+n)^2} \to^{m \to \infty} g(z)$ uniformly.
We define the function $G:K \setminus \Bbb{Z} \to \Bbb{C}$ as $G(z):=\sum_{|n|<N}\frac{1}{(z+n)^2}+g(z)$
Let $m>N$.
Then $$\sup_{z \in K \setminus \Bbb{Z}}| \sum_{|n| \leq m}\frac{1}{(z+n)^2}-G(z)|=\sup_{z \in K \setminus \Bbb{Z}}| \sum_{N \leq |n| \leq m}\frac{1}{(z+n)^2}-g(z)| \to^{m \to \infty}0$$
$2)$ $K$ does not contain any integers.
We do the same procedure.
Is this correct?
This question is from some old notes in complex analysis.
The same notes have been slightly modified and there is a different formulation of this problem,where we must prove that the series converges uniformply for every compact $K \subseteq \Bbb{C} \setminus \Bbb{Z}$(which is different from $K \setminus \Bbb{Z}$ since this set may not be compact at all)
Is my proof correct,or is the problem wrong and that is why was modified?
Thank you in advance for your feedback.