How to investigate the uniform convergence of the series

Question

Given this series: $$\sum_{n=1}^\infty \frac{1}{n(1+x^2)^n}, $$ for which values of $x$ is it uniformly convergent?

Determining the points for which it is point-wise convergent is very easy, by just using the root test, I came up with a result that the series is point-wise convergent for all $x\in\mathbb{R}\setminus\{0\}$.

However, I have no idea on how to get the values of $x$ for which this series is uniformly convergent. Am I supposed to use the definition for uniform convergence or the Weierstrass M test?

Answer 1

The series is uniformly convergent in any subset of $I_a:=(-\infty,-a]\cup [a,+\infty)$ with $a>0$. For $x\in I_a$, $$0<\frac{1}{n(1+x^2)^n}\leq \frac{1}{n(1+a^2)^n}$$ and $\sum_{n=1}^\infty \frac{1}{n(1+a^2)^n}$ is a convergent series (use the ratio test). Then uniform convergence follows from the Weierstrass M-test.

However, the series is not uniformly convergent in $\mathbb{R}\setminus\{0\}$. For $x\not=0$, let $f(x)=\sum_{n=1}^\infty \frac{1}{n(1+x^2)^n}$. Then $$\begin{align} \sup_{x\not=0}\left|f(x)-\sum_{n=1}^{N}\frac{1}{n(1+x^2)^n}\right| &=\sup_{x\not=0}\sum_{n=N+1}^{\infty}\frac{1}{n(1+x^2)^n}\\ &\geq \sum_{n=N+1}^{\infty}\frac{1}{n(1+1/n)^n} \quad\text{($x=1/\sqrt{n}$)}\\ &\geq \frac{1}{3}\sum_{n=N+1}^{\infty}\frac{1}{n}=+\infty \end{align}$$ where at the last line we applied $(1+1/n)^n\leq 3$ (or we may note that $(1+1/n)^n\to e$).

Answer 2

Take $a>0$. Then$$|x|\geqslant a\implies\frac1{1+x^2}\leqslant\frac1{1+a^2}\implies(\forall n\in\mathbb{N}):\frac1{n(1+x^2)^n}\leqslant\frac1{n(1+a^2)^n}$$and therefore the series converges uniformly on $\mathbb{R}\setminus(a,a)$, by the Weierstrass $M$-test.