Uniform Convergence of $\sum_{i=1}^\infty \arctan\left(\frac{x}{i^2}\right)$ and its differentiabilty

Question

I was trying to prove it is uniform convergent by it is Cauchy in sup-norm, since I don't know what does it converge to and it seems that M-test fail (as each term is bounded by $\pi/2$).

$\left\|\sum_{i=n+1}^m\arctan\left(\frac{x}{i^2}\right)\right\|$, then I dont know how to continue.

But I know it is pointwisely convergent.

Then, I differentiate it term by term, i get $\sum_{i=1}^\infty \frac{1}{\frac{x^2}{i^4}+1}$. Then I find that when $x=0$, it becomes $1+1+1+1+...$ which is divergent, but the question says $\sum_{i=1}^\infty \arctan\left(\frac{x}{i^2}\right)$ is differentiable.

Why does it happen? Thank you very much.

Answer 1

M-test works if you're working on a bounded interval - just need that $\vert \arctan x \vert \le \vert x \vert$ and that $\sum_i \frac{1}{i^2}$ is convergent.

A tidy proof of this for $x>0$ comes from the following:

$$0\le\frac{1}{1+t^2}\le1\implies0\le\int_0^x\frac{1}{1+t^2}dt\le\int_0^x1dt\implies0\le\arctan x\le x$$

and an entirely analogous proof works for $x<0$ as well.

If you're on an unbounded interval, the convergence isn't uniform - note that:

$$\sup_{x\in\mathbb{R}}\vert\sum_{n\ge N} \arctan \frac{x}{n^2}\vert\ge\arctan\frac{N^2}{N^2}=\frac{\pi}{4}$$

which shows that the sum is not Cauchy.

Furthermore, for your differentiation, you're missing a term from the chain rule:

$$\frac{d}{dx}\arctan\frac{x}{i^2}=\frac{1}{i^2}\frac{1}{1+\left(\frac{x}{i^2}\right)^2}=\frac{i^2}{x^2+i^4}$$

Answer 2

For starters, $$ \frac{d}{dx}\arctan\frac{x}{k^2} = \frac{\frac{1}{k^2}}{1+\left(\frac{x}{k^2}\right)^2} =\frac{1}{k^2+\frac{x^2}{k^2}}$$ and for $x\to 0$ $$ \sum_{k\geq 1}\frac{1}{k^2}=\frac{\pi^2}{6}.$$ Moreover, by the AM-GM inequality $$ \frac{1}{k^2+\frac{x^2}{k^2}}=\frac{1}{\frac{k^2}{2}+\frac{k^2}{2}+\frac{x^2}{k^2}}\leq \frac{2}{k^2+|x|\sqrt{8}} $$ and for any $A\in\mathbb{R}^+$: $$ \sum_{k\geq 1}\frac{1}{k^2+A^2}=\frac{-1+\pi A\coth(\pi A)}{2A^2}.$$ So it is not difficult to prove the pointwise convergence of the series of the derivatives to a continuous function, by studying it first on $[0,1]$ then on $(1,+\infty)$. Also notice that $$ \prod_{k\geq 1}\left(1+\frac{ix}{k^2}\right) \tag{1}$$ is a convergent product, since $\sum_{k\geq 1}\frac{x}{k^2}=\frac{\pi^2 x}{6}$, and $$ \sum_{k\ge 1}\arctan\left(\frac{x}{k^2}\right) $$ is trivially related with the argument of $(1)$.