Does the series converges uniformly on $\mathbb{R}$?

Question

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 $\mathbb{R}$?

Answer 1

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 $x\in \mathbb{R}$.

Answer 2

Set $$s_{n}(x):=\sum_{n=1}^{\infty }(-1)^n\frac{n}{n^2+x^2},\quad u_{n}(x):=\frac{n}{n^2+x^2},$$ $$s(x):=\lim_{n \to \infty} s_{n}(x).$$ $\\$

For any fixed $x_{0}\in\mathbb{R}$, the derivative of $u_{n}(x)$ with respect to $n$ is $f^{'}(n)=\frac{x^2_0-n^2}{(x^2_0+n^2)^2}.$ $f^{'}(n)<0,$ when $n\geq \left \lfloor\left | x_{0} \right | \right \rfloor +1.$ Next we fix $n\in \mathbb{N}^{\ast}$, for any $x\in [-n,n]$, and then have $$\left | r_{n}(x)\right |:=\left | s_{n}(x)-s(x) \right |=|\sum_{k=n+1}^{\infty }(-1)^k u_{k}(x)| \leq\frac{n+1}{(n+1)^2+x^2}\leq\frac{1}{n}.$$

$\\$

Let $a_{k}:=\sup_{x\in [-k,k]}|s_k(x)-s(x)|,$ $\sup_{k\geq n}\left \{ a_{k} \right \} \leq\frac{1}{n}$holds from the above statement. Under such circumstance,we can prove $$\lim_{n \to \infty} \sup_{k\geq n}\left \{ a_{k} \right \} =0 \quad (i.e.\limsup_{n \to \infty} a_{n}=0 )$$ is equivalent to $$\lim_{n \to \infty} \sup_{x\in\mathbb{R}} |s_{n}(x)-s(x)|=0.$$