<s>[INST] Show that the series converges uniformly Show that $\sum_{n=1}^{\infty} \sin \left(\dfrac{x}{n}\right)^2$ converges uniformly on $[a,-a]$  , $a\in \mathbb{R}$.
My attempt:
Using Taylor's formula, we have:$$ \sin\left(\dfrac{x}{n}\right)^2 = \dfrac{x^2}{n^2} + o\left( \dfrac{x^2}{n^2}\right)$$
Therefore, 
$$\sin\left(\dfrac{x^2}{n^2}\right) \leq \dfrac{x^2}{n^2} \leq \dfrac{a^2}{n^2}$$
We then have $\| \sin\dfrac{x^2}{n^2}\| \leq \dfrac{a^2}{n^2} \leq \infty$
Therefore, according to Weierstrass's criteria:$\sum_{n=1}^{\infty} \sin \left(\dfrac{x}{2}\right)^2$   converges uniformly on $[a,-a]$
Is that the right approach to the problem?
 [/INST] Since your $x$ is bounded, there exists an $N$ such that 
\begin{equation}
0<\frac{x^2}{n^2}\leq \frac{\pi}{2}
\end{equation}
for all $n\geq N$ and $x\in[-a,a]$. Now you can use the inequality $0\leq \sin x\leq x $ for $0\leq x\leq \frac{\pi}{2}$.
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