Uniform convergence of $\sum_{n=1}^{\infty}\frac{nx}{n^2+x^2}$

Question

Does the following series converge uniformly on $x \in [-1,1]$? What about $x \in \mathbb{R}$?

$$ \sum_{n=1}^{\infty}\frac{nx}{n^2+x^2} $$

My attempt:

Series converge pointwise to $$ f(x) = 0 $$ when $n \rightarrow \infty$.

I want to ensure that $$ \sup_{x \in E} \left| \frac{nx}{n^2+x^2} \right| \lt \epsilon $$

by calculating derivative we obtain: $$ f_n^{'}(x)=\frac{n(n^2-x^2)}{(n^2+x^2)^2} = 0 \iff x = \pm {n} $$ We also know that $f_n$ increases on $(-\infty, -{n})$, decreases on $(-{n}, {n}$) and then increases to infinity. That means $-{n}$ is its local maximum. Plugging that into our function: $$ f(-{n}) = \frac{-n^2}{2n^2} = -\frac{1}{2} $$ I am not sure what can I conclude from here. Is that enough for $\mathbb{R}$ not having uniform convergence? Any hints appreciated.

Answer 1

About the original task formulation: The series $$\sum_{n=1}^\infty\frac{nx}{n^2+x^2}=x\sum_{n=1}^\infty\frac{n}{n^2+x^2}$$ does not converge at all, obviating the question about uniform convergence, as for $n>|x|$ you get $$\frac{n}{n^2+x^2}>\frac1{2n}.$$ Thus the harmonic series $$\sum_{n=1}^\infty \frac1{2n}$$ is a diverging minorant, proving divergence.

Answer 2

I am choosing to interprete your question as the uniform convergence of $$f_n := \frac{nx}{n^2 + x^2},$$ because your question as stated does not line up with the rest of your argument.

Note that on the interval $[0,1]$, we have that $$f_n(x) = \frac{nx}{n^2 + x^2} \leq \frac{n}{n^2 + x^2} \leq \frac{n}{2n^2} \leq \frac{2}{n},$$ so we have uniform convergence.