Let $$f(z)=\sum_{n=0}^\infty \frac 1 {z+n^2}\ .$$ Prove that $f$ is holomorphic in $\mathbb C \setminus \{ -n^2:n\in \mathbb N\} $, and determine the type of the singularities in $\{ -n^2:n\in \mathbb N\}$.
It is clear that the series converges pointwise in any $z \in \mathbb C$, provided that every term of the sum is defined: namely, in $\mathbb C \setminus \{ -n^2:n\in \mathbb N\} $. As $z$ approaches $-N^2$ for any $N\in \mathbb N$, $f$ can be regarded as $$f(z)=\sum_{n=0,\ n \neq N}^\infty \Bigl(\frac 1 {z+n^2}\Bigl) + \frac 1 {z+N^2} \ .$$ The sum converges, while the other term has obviously a simple pole in $-N^2$.
However the pointwise convergence isn't enough to prove that $f$ is holomorphic; actually, I should prove that the series converges uniformly over any compact set $A\subset \mathbb C \setminus \{ -n^2:n\in \mathbb N\} $. For $n$ large enough, $\frac 1 {|z+n^2|}$ tends to the same value for every $z\in A$; and of course for every such $z$ the series $\sum_{n=0}^\infty \frac 1 {|z+n^2|}$ converges. Clearly my intention is to use Weierstrass criterion; however I don't understand whether this is a rigorous reasoning or it is too heuristic. Thanks in advance
You have indeed the right idea: try to apply Weierstrass criterion. Let $S=\{-n^2: n\in\mathbf{N}\}$ be your singularities and $K\subseteq \mathbf{C}\setminus S$ be any compact set. We want to upper bound $1/|z-n^2|$ for $z\in K$ by a quantity independent of $z$. Note that over a compact set, the norm of $z$ cannot be too large (let's say $|z|\le M$ for all $z\in K$), and therefore:
$$|z+n^2|\ge n^2-|z|\ge n^2-M$$
Now we can apply Weierstrass criterion starting at $n$ large enough (so that $n^2\ge M$ and we don't have problems with signs): $\sum_n 1/(n^2-M)$ converges so $\sum_n 1/|z+n^2|$ converges uniformly over compact sets $K\subseteq \mathbf{C}\setminus S$. This proves the first part of your question.
For the second part, what happens at the neighborhood of $-k^2$ if you multiply $f$ by $z+k^2$?