I am trying to prove that this infinite series $\sum_{n=1}^{\infty}\frac{x^{2}}{x^{2}+n^{2}}$ does not converge uniformly on $(-\infty,\infty)$.
I can definitely show that this series converges uniformly on $[-R,R]$ for any $R$ using Weierstrass $M$-test, and I understand that even though this uniform convergence holds for all $R\in\mathbb{R}$, it may still be not true that the uniform convergence on $(-\infty,\infty)$.
However, I have no doable idea to prove that the series does not converge uniformly on $(-\infty,\infty)$. My attempt is to show that $$\lim_{n\rightarrow\infty}\sup_{x\in(-\infty,\infty)}\Big|\sum_{k=1}^{n}\dfrac{x^{2}}{x^{2}+k^{2}}-\sum_{k=1}^{\infty}\dfrac{x^{2}}{x^{2}+k^{2}}\Big|=\lim_{n\rightarrow\infty}\sup_{x\in(-\infty,\infty)}\Big|\sum_{k=n+1}^{\infty}\dfrac{x^{2}}{x^{2}+k^{2}}\Big|\neq0.$$
The idea is then try to found a lower bound of $\frac{x^{2}}{x^{2}+k^{2}}$ for each $k$, and this lower bound is easy to be computed in the infinite sum. However, it is hard to find such a bound.
I firstly separate the domain to $x\geq 0$ ad $x\leq 0$. For example, for $x\geq 0$, $$x^{2}+k^{2}=(x+k)^{2}-2xk\leq (x+k)^{2},$$ and thus $$\dfrac{x^{2}}{x^{2}+k^{2}}\geq \dfrac{x^{2}}{(x+k)^{2}}=\Big(\dfrac{x}{x+k}\Big)^{2},$$ but then I still cannot cancel $x$ both top and bottom so that I have a convergence series but the sup in $x$ will blow up.
Is there any other way to approach this question? Thank you!
Best Answer
Note that if $\sum_{n=1}^\infty f_n$ converges uniformly in some domain $D$ then $f_n$ converges uniformly to $0$ in $D$. The proof is very similar to the standard proof that if a series of numbers converges then its general term tends to $0$.
So it is enough to show the sequence of functions $f_n(x)=\frac{x^2}{x^2+n^2}$ doesn't converge uniformly to $0$ in $\mathbb{R}$. We have to show that there is some $\epsilon>0$ such that for all $n_0\in\mathbb{N}$ there are $n\geq n_0$ and $x\in\mathbb{R}$ such that $|f_n(x)-0|\geq\epsilon$. Indeed, let $\epsilon=\frac{1}{2}$. Then for each $n_0\in\mathbb{N}$ you can take $n=n_0$ and $x=n_0$. They satisfy $|f_n(x)-0|=\frac{n_0^2}{n_0^2+n_0^2}=\frac{1}{2}\geq\epsilon$.