Convergence of $sum_{n=1}^infty frac{(-1)^n}{x^2-n^2}$ in the reals and in the complex numbers

Just for your curiosity. $$sum_{n=1}^infty frac{(-1)^n}{x^2-k n^2}=frac{pi x csc left(frac{pi }{sqrt{k}}xright)-sqrt{k}}{2 sqrt{k}, x^2}$$ which is defined in $mathbb{R}$ and $mathbb{C}$

If $;z=alpha+ibeta;$ 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 $zinmathbb{C}$.

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 $a_n$.

$$left|a_nright|=left|frac{(-1)^n}{x^2-n^2}right|=left|frac{1}{n^2-x^2}right|.$$

$x$ cannot be a non-zero integer, otherwise some $a_n$ is not defined, so $xinleft(mathbb{C}setminusmathbb{Z}right)cup{0}$.

Moreover, for all $ninmathbb{N}$ such that $n>2|x|$ we get $$left|a_nright|=left|frac{1}{n^2-x^2}right|lefrac{1}{left|n^2right|-left|x^2right|}=frac{1}{n^2-left|xright|^2}<frac{1}{n^2-frac{n^2}{4}}=frac{4}{3n^2}$$ 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 $xinleft(mathbb{C}setminusmathbb{Z}right)cup{0}$.