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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?

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    $\begingroup$ It should be uniformly convergent on any subset away from $0$, i.e. $|x| > \epsilon$ for some $\epsilon > 0$. $\endgroup$ – mathworker21 Feb 25 '18 at 8:38
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    $\begingroup$ Uniform convergence only makes sense on sets of points, not at individual values. $\endgroup$ – Lord Shark the Unknown Feb 25 '18 at 8:38
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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 $f(x)=\sum_{n=1}^\infty \frac{1}{n(1+a^2)^n}$. Then $$\begin{align} \sup_{x\not=0}\left|f(x)-\sum_{n=1}^{N}\frac{1}{n(1+x^2)^n}\right| &=\sup_{x\not=0}\sum_{n=N+1}^{\infty}\frac{1}{n(1+x^2)^n}\\ &\geq \sum_{n=N+1}^{\infty}\frac{1}{n(1+1/N)^n} \quad\text{($x=1/\sqrt{N}$)}\\ &\geq \sum_{n=N+1}^{\infty}\frac{1}{N(1+1/N)^N}=+\infty \end{align}$$ because $N(1+1/N)^N\sim Ne.$

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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.

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