I face a problem now, it is known that $f_n(x)$ converge to $+\infty$ pointwise as $n\rightarrow+\infty$, but I want to prove $f_n(x)$ converge to $+\infty$ uniformly.
I can prove $\arctan f_n(x)$ converges to $\frac{\pi}{2}$ uniformly. Does this imply that $f_n(x)$ converge to $+\infty$ uniformly?
On
Assume we have a sequence of functions $f_n(x)$ defined on some set $D\subset \mathbb{R}$ such that $\arctan f_n(x) \to \pi/2$ uniformly on $D$. We prove that $f_n $ converge to $+\infty$ uniformly, i.e. $$ \forall A > 0 \ \ \exists N = N(A) \in \mathbb{N} \ \ \ \text{such that if } \ \ n>N \ \ \text{ then } \ f_n(x) > A \ \ \forall x \in D. $$ Assume for contradiction, that the above does not hold, that is $$ \tag{1} \exists A > 0 \ \ \text{s.t.} \ \ \forall N \in \mathbb{N} \ \ \exists x_N \in D \ \ \text{s.t.} \ \ f_N(x_N) \leq A . $$
In view of uniform convergence of $\arctan f_n$ we have that for any $\varepsilon > 0$ there is $N = N(\varepsilon) \in \mathbb{N}$ such that whenever $n> N$ then $|\arctan f_n(x) - \pi/2| < \varepsilon$ for all $x\in D$ . Combining this with $(1)$ and using the fact that $\arctan$ is increasing, from $(1)$ we get $$ \pi/2 - \varepsilon < \arctan f_n(x_n) \leq \arctan A < \pi/2 \tag{2} $$ if $n>N(\varepsilon)$. Since $A>0$ is fixed, we can choose $\varepsilon> 0$ small enough to make the inequality $(2)$ false. This contradiction proves uniform convergence of $f_n$ to $\pi/2$ on $D$.
Yes it does, because $\tan$ is a bijective and increasing mapping from $[0, \frac \pi 2)$ to $[0, \infty) $.
More detailed: Given $M > 0$ we have that $0 < \arctan M < \frac \pi 2$. The uniform convergence of $\arctan f_n(x) \to \frac \pi 2$ implies the existence of an $n_0 \in \Bbb N$ such that $$ \arctan f_n(x) > \arctan M \text{ for all $n \ge n_0$ and all $x$.} $$ It follows that $$ f_n(x) > M \text{ for all $n \ge n_0$ and all $x$.} $$