Let $f_{n}(x) = \frac{\arctan(nx)}{\sqrt{n}}$ show that it converges uniformly to $f$ - Proof Verification

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Let $f_{n}(x) = \frac{\arctan(nx)}{\sqrt{n}}$, show that it converges uniformly to $f$ .

Examining the function, we see that $f_{n} \rightarrow 0$ as we take $n \rightarrow \infty$

So it becomes having to show:

$$\|f_{n}(x) - f(x) \|_{\infty} \rightarrow 0$$

Therefore:

$$\|f_{n}(x) - f(x) \|_{\infty} = \sup_{x \in \mathbb{R}}\Bigg|\frac{\arctan(nx)}{\sqrt{n}}\Bigg| < \frac{\frac{\pi}{2}}{\sqrt{n}} \\ \Rightarrow \lim_{n \rightarrow \infty} \frac{\frac{\pi}{2}}{\sqrt{n}} \rightarrow 0 $$

So this is the scratch work. Now how do I formally state to take a specific $n > N$ ? I'm trying to tie together my ability to write out this process in a good mathematical manner.

Thanks in advance.