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Verify the following series converges uniformly on $\mathbb{R}$ $$\sum_{n=1}^{\infty}(-1)^n\frac{n}{n^2+x^2}$$
$\forall M>0,$ we have that the series is converges uniformly on $[-M,M]$ by using The Dirichlet test for uniform convergence. How can I prove it converges uniformly on $\mathbb{R}$?

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  • $\begingroup$ $\frac{n}{n^2+x^2}$ converges monotonically depend on $x$. When $\sqrt{n^2+n} \leq x \leq \sqrt{n^2+n},u_{n}(x)>u_{n+1}(x).$ NOT for all $x\in \mathbf{R}!$ $\endgroup$ – Elliot Oct 19 '17 at 13:18
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I try an answer.

We have $$\frac{n}{n^2+x^2}=\frac{1}{n}-\frac{x^2}{n(n^2+x^2)}=\frac{1}{n}-v_n(x)$$ To prove the uniform convergence on $\mathbb{R}$, it suffice to show the uniform convergence of the series $(-1)^nv_n(x)$ on $\mathbb{R}$, as we have for an $N$ $$|\sum_{n\geq N}(-1)^nu_n(x)|\leq |\sum_{n\geq N}\frac{(-1)^n}{n}|+|\sum_{n\geq N}(-1)^nv_n(x)|$$ But now, the sequence $v_n(x)$ is decreasing and $\to 0$ for all $x\in \mathbb{R}$, and we can use the usual criterion.

More precisely, we have $|\sum_{n\geq N}\frac{(-1)^n}{n}|\leq \frac{1}{N}$ and $|\sum_{n\geq N}(-1)^nv_n(x)|\leq v_N(x)\leq \frac{1}{N}$, and so $|\sum_{n\geq N}(-1)^nu_n(x)|\leq \frac{2}{N}$ for all $N\geq 1$ and all $x\in \mathbb{R}$.

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  • $\begingroup$ very fancy answer $\endgroup$ – Guy Fsone Oct 19 '17 at 15:05

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