Uniform convergence $f_n(x)=\arctan (nx)$. Assume $f_n(x)=\arctan (nx)$.
Determining an interval of uniform convergence of the succession. In the interval $(-\infty, a)$, with a<0, the sequence converges uniformly?
Thank you very much
 A: For $a\leq0, f_{n}$ converges pointwise to the function $f(x)=-\frac{\pi}{2}$.
What you must observe when evaluating $$\lim_{n \to \infty} \sup_{x \in (-\infty, a)} |arctg(nx)+\frac{\pi}{2}|$$ is that $ \frac{\pi}{2}-arctg(x)=arctg(\frac{1}{x})$ So you get: $$|arctg(nx)+\frac{\pi}{2}|=|\frac{\pi}{2}-arctg(-nx)|=|arctg (-\frac{1}{nx})|=arctg (-\frac{1}{nx})$$ since $-\frac{1}{nx}\geq 0$. But the arctg is a monotonically increasing function, so: $$\sup arctg (-\frac{1}{nx})= arctg (-\frac{1}{na}) \rightarrow 0,$$ for $n\rightarrow \infty.$ Which proves uniform convergence.
A: First, observe that the function $f(x)=-\frac{\pi}{2}$ is the pointwise limit of this sequence on $(-\infty,a)$.
Now, we prove that the sequence $f_n(x)=\arctan(nx)$ converges uniformly to $f(x)$ on $(-\infty,a)$ as $n\to\infty$. Since the function $\arctan(nx)$ increases with $x$, it reaches its supremum (least upper bound) on the interval $(-\infty,a)$ at $x=a$ for all $n$. Moreover, it is strictly negative on this interval and $\displaystyle{\lim_{x\to-\infty}}\arctan(nx)=-\frac{\pi}{2}$. We see that
$$0<|\arctan(a)|\le|\arctan(na)|\le|\arctan(nx)|\le|-\frac{\pi}{2}|\,\,\text{for all }x.$$
Furthermore, since $f(x)=-\frac{\pi}{2}$, it follows that
$$0\le|f(x)-\arctan(nx)|\le|-\frac{\pi}{2}-\arctan(na)|.$$
Now, note that $\displaystyle{\lim_{n\to\infty}}\arctan(na)=-\frac{\pi}{2}$. Consequently, for every $\varepsilon>0$, there exists an $N$ such that for all $n>N$, it is true that $|-\frac{\pi}{2}-\arctan(na)|<\varepsilon$. It follows that for all $x\in(-\infty,a)$, we have that
$$|f(x)-f_n(x)|=|f(x)-\arctan(nx)|\le|-\frac{\pi}{2}-\arctan(na)|<\varepsilon,$$
so we are done.
