Is $\sum\limits_{n=1}^\infty \arctan \frac x {n^2}$ continuous?

Question

Let $$ f(x) = \sum\limits_{n=1}^\infty \arctan \frac x {n^2} $$ I need to check whether $f : \mathbb R\to \mathbb R$ is continuous.

Of course, if it converges, $f(x) = -f(-x)$, so I will be only concerned about nonnegative $x$ (the series terms are in this case nonnegative).

Of course, $f$ converges pointwise, as for every $x\geq 0$, $\arctan x\leq x$. In particular, from comparative test, $$ \sum\limits_{n=1}^\infty \arctan \frac x {n^2} \leq x \sum\limits_{n=1}^\infty \frac 1 {n^2} < x\cdot\infty=\infty. $$

To see if it's continuous, we could check if it converges uniformly. However, if $\sum\limits_{n=1}^\infty \arctan \frac x {n^2}$ converges uniformly, then $\arctan \frac x {n^2}$ converges uniformly to $0$. The last one is false though, as $||\arctan \frac x {n^2}|| = \sup_{x\geq 0}|\arctan \frac x {n^2}| = \pi/2 \not \to 0$.

I don't know any theorem about continuity of non-uniform convergence of series though. How can it be checked?

Answer 1

You can work locally: so let $a>0$ and then

$$\Vert\arctan \frac{x}{n^2}\Vert=\sup_{0\le x\le a} \vert \arctan\frac x{n^2} \vert\le \arctan \frac{a}{n^2}\sim \frac a{n^2}$$ so the series is uniformly convergent on every interval $[0,a]$ so $f$ is continuous on $[0,+\infty)=\bigcup\limits_{a>0}[0,a]$.

Answer 2

hint

To prove that the odd function $f$ is continuous at $\Bbb R$, you just need to prove it is continuous at $[0,A]$ for all $A>0$.

use your inequality to get $$|\arctan(\frac{x}{n^2})|\le \frac{A}{n^2}$$

Answer 3

For $A >0$, take a compact $[-A,A] \subseteq \mathbb R$.

using the inequality you proved, you have for $x \in [-A,A]$: $$\sum_{n=1}^\infty \arctan \frac x {n^2} \leq x \sum_{n=1}^\infty \frac 1 {n^2} < \vert x \vert \sum_{n=1}^\infty \frac 1 {n^2} \le A \sum_{n=1}^\infty \frac 1 {n^2} .$$

And the RHS is finite as $\sum_{n=1}^\infty \frac 1 {n^2}$ is a convergent series. Therefore the series converges uniformly on all compacts and is is continuous on all compacts and therefore continuous on $\mathbb R$.

Answer 4

The inverse tangent has a bounded derivative $$ \left\lvert\frac{d}{dx} \arctan x \right\rvert = \frac{1}{1+x^2} \le 1 $$ and is therefore Lipschitz continuous: $$ |\arctan x - \arctan y | \le |x-y| $$ for all $x,y \in \Bbb R$. It follows that the partial sums satisfy $$ \left\lvert \bigl( \sum_{n=1}^N \arctan\frac x {n^2}\bigr) - \bigl(\sum_{n=1}^N\arctan \frac y {n^2}\bigr) \right\rvert \le |x-y|\sum_{n=1}^N \frac{1}{n^2} $$ for all $N$. Now take the limit $N \to \infty$ to conclude $$ |f(x) - f(y) | \le \frac{\pi^2}{6} |x-y| $$ so that $f$ is Lipschitz (and thus uniformly) continuous on $\Bbb R$.