Check if $f(x)=\sum_{n=1}^{\infty}\frac{1}{x^2-n^2}$ is continuous and differentiable function $$ D = \mathbb{R} \setminus \mathbb{Z}$$
My try
$$\frac{1}{x^2-n^2} \text{~~~} 1/n^2$$ so $$\sum_{n=1}^{\infty}\frac{1}{x^2-n^2}$$ converges (pointwise) as well $$ \sup_{x \in D} \left| \frac{1}{n^2 - x^2} - 0 \right| = \sup_{x \in D} \left| \frac{1}{n^2 - x^2} \right| = \infty $$ but I don't know how to continue that way..
If $x$ is not an integer then the inequality $|x^{2}-n^{2}| \geq n^{2}-x^{2}$ shows that the series is uniformly convergent in some neighborhood of $x$ and hence the sum is continuous at $x$. Apply the same argument to the differentiated series to conclude that the sum is in fact differentiable at any point which is not an integer as well as at $0$.