real analysis - Convergence of $sum_limits{n=1}^{infty} (-1)^n sin(frac{x}{n})$

Show that $\displaystyle\sum_{n=1}^{\infty} (-1)^n \sin \left(\frac{x}{n}\right)$ converges uniformly on every finite interval.

At first thought, I try to find a bound, $M_n$, for the sine term such that $\displaystyle\sum_{n=1}^{\infty} M_n < \infty$.

However, the best I can find is that $\sin(\frac x n) \leq \frac R n + \frac{1}{n^3}$ for $x \in [-R,R]$

And now, I totally don't know how to proceed.

Thanks in advance.

Answer

Hint. One may recall that, by a Taylor series expansion, as $u \to 0$, we have
$$\sin u= u+O(u^3)$$ giving, as $n \to \infty$,
$$\sin \frac{x}{n}= \frac{x}{n}+O_x\left(\frac{1}{n^3}\right)$$ and, for any $N\ge1, M \ge1,$
$$\sum_{n=N}^M(-1)^n\sin \frac{x}{n}=x \cdot \sum_{n=N}^M\frac{(-1)^n}{n}+\sum_{n=N}^M O_x\left(\frac{1}{n^3}\right)$$ one may deduce that

$$\left|\sum_{n=N}^M(-1)^n\sin \frac{x}{n}\right|\le |x| \cdot\left|\sum_{n=N}^M\frac{(-1)^n}{n}\right|+\sum_{n=N}^M\left|O_x\left(\frac{1}{n^3}\right)\right|$$ which yields the uniform convergence of the given series over each compact set $[-R,R]$, $R>0$.