Consider the derivative series
$$ D = \sum_{n=1}^\infty \frac{-2x}{(2+x^2)^{n+1}} $$ of $$ S = \sum_{n=1}^\infty \frac{1}{n} \frac{1}{(2+x^2)^n} $$
I have shown that $S$ converges by the use of Weiterstrass' M-test by saying that as $x \in \mathbb{R}$ we must have that $$ \frac{1}{(2+x^2)^n} \leq \frac{1}{n} $$ for all $n \in \mathbb{N}$. This means that $$ \frac{1}{n} \frac{1}{(2+x^2)^n} \leq \frac{1}{n^2} $$ and from analysis we know that $$ \sum_{n=1}^\infty \frac{1}{n^2} $$ converges. But I am struggling to prove that $D$ converges uniformly. Do I have to use the same approach? I found that for $x \in \mathbb{R}$ and for all $n \in \mathbb{N}$ that $$ \frac{-2x}{(2+x^2)^{n+1}} \leq \frac{1}{2^n} $$ which is a geometric series that converges. Can I use this? Thanks in advance.
If $f(x)=\frac{-2x}{(2+x^2)^{n+1}}$, then$$f'(x)=2 \left(x^2+2\right)^{-n-2} \left((2 n+1) x^2-2\right).$$So, the maximum of $|f|$ is attained when $x=\pm\sqrt{\frac2{2n+1}}$ and$$\left|f\left(\pm\sqrt{\frac2{2n+1}}\right)\right|=2 \sqrt{2} \sqrt{\frac{1}{2 n+1}} \left(\frac{2}{2 n+1}+2\right)^{-n-1}.$$Since$$\lim_{n\to\infty}2 \sqrt{2} \sqrt{\frac{1}{2 n+1}} \left(\frac{2}{2 n+1}+2\right)^{-n-1}=0,$$the convergence is uniform by the Weierstrass $M$-test.