Does $f_{n}\left(x\right)=x\arctan\left(nx\right)$ converge uniformly? Does $f_{n}\left(x\right)=x\arctan\left(nx\right)$ converge uniformly when $x\in\left(0,\infty\right)$?
I know that the limit function is $f\left(x\right)=\frac{\pi}{2}x$.
I think it does converge uniformly but I couldn't prove it.
I tried Cauchy criterion, I tried by definition and I tried to find:
$\lim_{n\rightarrow\infty}\sup_{x>0}\left|x\arctan\left(nx\right)-\frac{\pi}{2}x\right|$ 
with no success.
 A: Hint  Let $f_n(x)=\tan^{-1}nx$. If $m\geq n$, then $f_m(x)\geq f_n(x)$, and these functions are all positive. Note that proving that $f_n\to f$ uniformly is equivalent to showing $|f-f_n|\to 0$ uniformly, so consider the functions $$g_n(x)=|x|\left|\frac {\pi} 2-f_n(x)\right|$$ for $x>0$. We can see that $g_n\to 0$ pointwise over $[0,\infty)$. Moreover, $g_{n+1}(x)\leq g_n(x)$ for any $x\in[0,\infty)$. Now, you can split the proof in two parts. First, choose a large $M$ to make $$\left|\frac{\pi}2-f_n(x)\right|$$ small for any $x\geq M$. Then split over $[0,M]$ and $[M,\infty)$. On $[0,M]$ use Dini's Theorem, and for $[M,\infty)$ you can use the fact that 
$$ \frac{π}   2 - \tan^{-1}(n x)=\tan^{-1}\frac{1}{nx}=\frac{1}{nx}-\frac{1}{3n^2x^2}+O\left(\frac1{n^3x^3}\right)$$
A: Using that $\arctan x=\int_0^xdt/(1+t^2)$ we have for $x>0$
$$
\Bigl|\frac{\pi}2\,x-x\,\arctan(n\,x)\Bigr|=x\int_{nx}^\infty\frac{dt}{1+t^2}=\frac1n(n\,x)\int_{nx}^\infty\frac{dt}{1+t^2}\le\frac1n.
$$
The last inequality holds because for any $z>0$
$$
z\int_z^\infty\frac{dt}{1+t^2}\le z\int_z^\infty\frac{dt}{t^2}=1.
$$
