Taking the derivative of $\sum_\limits{n=1}^{\infty}\arctan(\frac{x}{n^2})$

Question

Problem: Study the possibility of taking the derivative of the following series:

$$\sum_\limits{n=1}^{\infty}\arctan\left(\frac{x}{n^2}\right)\:\:,x\in\mathbb{R}$$

I have studied the following theorem:

Theorem: Suppose that $\sum_\limits{n=k}^{\infty}f_n$ converges uniformly to $F$ on $S=[a,b]$. Assume that $F$ and $f_n\:\:,n\geqslant k$, are integrable on $[a,b]$. Then:

$$\int_\limits{a}^{b}F(x)dx=\sum_\limits{n=k}^{\infty}\int_\limits{a}^{b}f_n(x)dx$$

Following the theorem I would need to check out if the derivative converges uniformly $\sum_\limits{n=1}^{\infty}(\frac{1}{1+\frac{x^2}{n^4}}\frac{2x}{n^4})$ I tried to apply Dirichlet to latter. Once I know that $\sum_\limits{n=1}^{\infty}\frac{2x}{n^4}$ by the integral test converges uniformly. However I was not able to apply it due to the fact that I could not prove $$\sum_\limits{n=1}^{\infty}\left(\frac{1}{1+\frac{x^2}{n^4}}\right)\leqslant M$$

Question:

How should I solve the problem?

How should I prove the series $\sum_\limits{n=1}^{\infty}\arctan\left(\frac{x}{n^2}\right)$ converge?

Thanks in advance!

Answer 1

Note that, if $f_n(x)=\arctan\left(\frac x{n^2}\right)$, then$$f_n'(x)=\frac{n^2}{n^4+x^2}\leqslant\frac{n^2}{n^4}=\frac1{n^2}.$$Besides, the series $\sum_{n=1}^\infty\frac1{n^2}$ converges.

Answer 2

The standard theorem is this:

If $\sum_{n=1}^{\infty}f_n(x)$ converges at least at one point, and the series of derivatives $\sum_{n=1}^{\infty}f_n^{'}(x)$ converges uniformly in some interval $I$, then in that interval the series can be differentiated term by term, meaning that the sum of the derivatives converges to the derivative of the sum.

In our case, the series clearly converges for all $x$. Moreover, the series of derivatives is: $$\sum_{n=1}^{\infty}\frac{n^2}{n^4+x^2}$$ which converges uniformly in $\mathbb{R}$ because for every $x\in\mathbb{R}$, the general term is bounded by $1/n^2$. It follows that the original series can be differentiated term by term.