Uniform convergence of $f_n(x) = n \sin(\frac{\pi x}{n})$ in [a,b]

Question

Show that the sequence of functions $$f_n(x) = n \sin \Bigl(\frac{\pi x}{n} \Bigr)$$

Converges uniformly in any interval of the form $[a,b]$

My attempt:

The pointwise limit of this function is $$ f(x) = \pi x$$

So I've tried to analyze (calculating the derivative, for example)

$$\Bigl| n sin \Bigl(\frac{\pi x}{n} \Bigr) - \pi x \Bigr| $$

without much success. Any hint?

Thanks in advance!

Answer 1

If follows from Taylor's Theorem that for all $y \in \mathbb R$, $$ \lvert \sin(y) - y \rvert \le \frac {\lvert y \rvert^3}{3!} $$

(compare estimating the error of $\sin(x) = x$ with Taylor's Theorem). Therefore

$$ \lvert n \sin(\frac{\pi x}{n}) - \pi x \rvert \le \frac {\pi^3 \lvert x \rvert^3}{6 n^2} \le \frac {\pi^3}{6 n^2} \cdot \max(\lvert a \rvert, \lvert b \rvert)^3 $$

for $x \in [a, b]$.