Test uniform convergence of the series
$$\sum_{n=1}^{\infty}\frac{x}{(n+x^2)^2}$$
My attempt:
Here $$f_n(x)=\frac{x}{(n+x^2)^2}$$
Applying Weierstrass's M-test
$$\vert \frac{x}{(n+x^2)^2}\vert < \frac{x}{n^2}<\frac{1}{n^2} \forall x\in(-1,1)$$
Since we know $\frac{1}{n^2}$ is convergent, the series converges for $x\in(-1,1)$
My question is how can I make sure that this is the only interval where the series is Uniformly convergent? And if it isn't, how do I find other such intervals?
On
Let $M$ be a bounded set in $ \mathbb R$. Then there is $c>0$ such that $|x| \le c$ for all $x \in M$. Hence:
$\vert \frac{x}{(n+x^2)^2}\vert \le \frac{|x|}{n^2}<\frac{c}{n^2} \forall x\in M$.
Therefore the seies is uniformly convergent on $M$.
On
We have, $$|\sum_{n=1}^{\infty} \frac{x}{(n+x^2)^2}|\le C\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}<\infty $$
Indeed we let $f_n(x)=\frac{x}{(n+x^2)^2}$ then $$f'_n(x)=\frac{n-3x^2}{(n+x^2)^3} = 0 \Longleftrightarrow x= \pm a_n =\pm\sqrt{\frac{n}{3}}$$
For $x\in (-\infty, -a_n)\cup (a_n, \infty)$ we have $f_n'\le 0$ then $f_n$ is nonincreasing on each interval. Namely,
$$-f_n(a_n)= f_n(-a_n) \le f_n(x)\le f_n(-\infty) =0 \qquad\text{for every $x\in (-\infty, -a_n)$}$$
$$f_n(a_n) \ge f_n(x)\ge f_n(\infty) =0 \qquad\text{for every $x\in (a_n, \infty)$}$$ in any case we have $$|f_n(x)|\le f_n(a_n) \qquad\text{for every $x\in (-\infty, -a_n)\cup (a_n, \infty)$}$$
and for $x\in (-a_n,a_n)$ we have $f_n'\ge 0$ so $f_n$ is nondecreasing on $(-a_n,a_n)$ i.e
$$-f_n(a_n)= f_n(-a_n) \le f_n(x)\le f_n(a_n)\qquad\text{for every $x\in (-a_n, a_n)$}.$$ Therefore for every $x\in\mathbb{R}$ , $$|f_n(x)|\le f_n(a_n) \approx \frac{1}{n^{3/2}}$$
With $|x|=y\sqrt{n}$, we have $$\frac{|x|}{(n+x^2)^2}=\frac1{n^{3/2}}\frac{y}{(1+y^2)^2}\le\frac{C}{n^{3/2}}$$ with some positive constant $C$, since $y/(1+y^2)^2$ is continuous for $y>0$ and $\rightarrow0$ as $y\rightarrow\infty$, so it must be bounded. Thus the series is uniformly convergent on the whole real line.