Computing derivatives of the sum $\sum_{n = -\infty}^\infty \frac{\lvert x \rvert}{(1 + n^2 x^2)^{3/2}}$

Question

Given a sum such as $\sum_{n = -\infty}^\infty \frac{\lvert x \rvert}{(1 + n^2 x^2)^\frac{3}{2}}$, the first derivative (with respect to $x$) of the summand $\frac{\lvert x \rvert}{(1 + n^2 x^2)^\frac{3}{2}}$ does not exist. Nonetheless, numerical estimates such as: seem to indicate that the sum itself is a smooth function of $x$ (and it seems like all derivatives vanish at $x=0$).

What tools are there to show that this sum defines a smooth function of $x$, and to compute the derivatives (especially at $x = 0$)?

Answer 1

The function is not even continuous at $x=0$: We have $f(0) = 0$, but for $x = \pm \frac 1k$ with $k \in \Bbb N$ is $$ f(x) \ge \lvert x \rvert\sum_{n = -k}^k\frac{1}{(1 + n^2 x^2)^\frac{3}{2}} \ge \frac 1k \frac{(2k+1)}{2^{3/2}} \ge \frac{1}{\sqrt 2} \, . $$

Answer 2

The limit as $x\to 0$ is basically the limit of the Riemann sum for $$\int_{-\infty}^\infty\frac{dt}{(1+t^2)^{3/2}}=\left.\frac{t}{\sqrt{1+t^2}}\right|_{-\infty}^\infty=2.$$ So the function has a discontinuity at $x=0$. On the other hand, the sum converges uniformly out of any neighborhood of $0$ (which is easy to show), hence it is continuous there (and the same argument applies for the smoothness).