Convergence of $\sum_{n=1}^\infty \frac{(-1)^n}{x^2-n^2}$ in the reals and in the complex numbers The goal is to find the values of $x\in\mathbb{R}$ (or $z\in\mathbb{C}$) such that the series
$$
\sum_{n=1}^\infty \frac{(-1)^n}{x^2-n^2}
$$
converges. The symbols $\mathbb{R},\mathbb{C}$ exclude $\lbrace -\infty,+\infty \rbrace$.
For the reals, since $x$ is finite, eventually $x<N\leq n$ for some $N\in \mathbb{N}$. Then the series converges provided that $x$ is not an integer. If I replace $x\rightarrow z$, I reach a similar conclusion.
Is the reasoning wrong and/or I omitted some special cases?
 A: If $\;z=\alpha+i\beta\;$ is any complex number then the non-negative square root of $\alpha^2+\beta^2$ is called the modulus of $z$ and it is denoted by $|z|$.
So by definition $\;|z|=\sqrt{\alpha^2+\beta^2}\;$ for all $z\in\mathbb{C}$.
In the particular case that $z$ is a real number, $|z|$ is equal to the absolute value of the real number.
We denote the elements of the series by $a_n$.
$$\left|a_n\right|=\left|\frac{(-1)^n}{x^2-n^2}\right|=\left|\frac{1}{n^2-x^2}\right|.$$
$x$ cannot be a non-zero integer, otherwise some $a_n$ is not defined, so $x\in\left(\mathbb{C}\setminus\mathbb{Z}\right)\cup\{0\}$.
Moreover, for all $n\in\mathbb{N}$ such that $n>2|x|$ we get $$\left|a_n\right|=\left|\frac{1}{n^2-x^2}\right|\le\frac{1}{\left|n^2\right|-\left|x^2\right|}=\frac{1}{n^2-\left|x\right|^2}<\frac{1}{n^2-\frac{n^2}{4}}=\frac{4}{3n^2}$$ and, by applying comparison test, it follows that
$$\sum_\limits{n=1}^\infty a_n=\sum_\limits{n=1}^\infty \frac{(-1)^n}{x^2-n^2}$$ is absolutely convergent for any $x\in\left(\mathbb{C}\setminus\mathbb{Z}\right)\cup\{0\}$.
A: Just for your curiosity.
$$\sum_{n=1}^\infty \frac{(-1)^n}{x^2-k n^2}=\frac{\pi  x \csc \left(\frac{\pi  }{\sqrt{k}}x\right)-\sqrt{k}}{2 \sqrt{k}\, x^2}$$ which is defined in $\mathbb{R}$ and $\mathbb{C}$
