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

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 $$x = 0$$)?

• You definitely have $f(0) = 0$, so your graph cannot be correct for small $x$. Sep 8, 2021 at 7:46
• FWIW, you can check that the limit at as $x\to 0$ of your function is equal to $\int_{-\infty}^{\infty} \frac{dy}{(1+y^2)^{3/2}}>0$ as an application of the integral test. Sep 8, 2021 at 8:26
• @Martin R There is no mistake, there is just a discontinuity at $x=0$. Sep 8, 2021 at 8:39
• @KeeleyHoek: What I meant is this: The graph indicates that $f(0) = 2$ and all derivatives vanish at $x=0$. But that is not correct because $f(0) = 0$. And if a function is not continuous then it has no derivatives at that point. Sep 8, 2021 at 8:43
• @metamorphy That is not correct. One sees directly at least for large $\lvert x \rvert$ that only the $n=0$ term survives and the sum is asymptotically $\lvert x \rvert$ , which obviously increases. Perhaps you were fooled by errors due to numerical precision, since the graph is very flat around $x=0$. Sep 8, 2021 at 8:43

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 $$f(x) \ge \lvert x \rvert\sum_{n = -k}^k\frac{1}{(1 + n^2 x^2)^\frac{3}{2}} \ge \frac 1k \frac{(2k+1)}{2^{3/2}} \ge \frac{1}{\sqrt 2} \, .$$
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).