Counterexample for functional sequence $\{f_n\}$ that converges uniformly on $E=[0;A]$ but not on $E=[0;+\infty)$

Question

Let functional sequence $\{f_n\}$ converges uniformly on $E=[0;A]$ $\forall A>0$. Does it implies this $[f_n]$ converges uniformly on $E=[0;+\infty)$?
My attempt: $$f_n(x)=\arctan nx$$

\begin{equation*} f(x) = \begin{cases} -\frac{\pi}{2} &\text{x<0}\\ 0 &\text{x=0} \\ \frac{\pi}{2} &\text{x>0} \end{cases} \end{equation*} Hence $f(x)=\arctan x \cdot \operatorname{sign} x$ $\Rightarrow$ if we take $x=\frac{1}{n} $ then $\sup r_n= \sup |\arctan 1 - \arctan\frac{1}{n}|$ $\Rightarrow$ $\lim\limits_{n\rightarrow +\infty}\sup r_n=\frac{\pi}{4}\not = 0$ hence $f_n(x)$ doesn't converge uniformly to $f(x)$. Am I right? Is there simpler counterexamples?

Answer 1

Consider $$ f_n(x)=\sum_{k=0}^n \frac{x^k}{k!} $$ Then $$ f_n(x)\to \mathrm{e}^x, $$ uniformly is EVERY closed interval, while it does not converge uniformly in $[0,\infty)$.

Answer 2

Here's yet another counterexample: $$f_n(x)=\min\{|x-n|,1\}.$$

Answer 3

$$f_n = 1_{[n, n+1]}\mbox{}\mbox{}$$