How do I show that $s=\sum\limits_{-\infty}^{\infty} {1\over (x-n)^2}$ on $x\not\in \mathbb Z$ is differentiable without using its compact form? I realize that the sequence of sums $s_a=\sum\limits_{-a}^{a} {1\over (x-n)^2}$ is not uniformly convergent.
I also tried to prove that it is continuous by using the usual $\varepsilon \over 3$ method. And it seems to apply because each $s_a$ is continuous and they converge pointwise to $s$. But then I realized that this should not be the right proof because I didn't use any special property of the given functions and the general case only works for uniform convergence. I am very confused. Please help!
Actually I do realize that if I could prove differentiability, continuity follows.
Thanks.