Convergence of $\sum_{n=1}^{\infty} (-1)^n \sin(\frac{x}{n})$ 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.
 A: 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 $[-R,R]$, $R>0$.
A: Let $b>0.$ We'll show uniform convergence on $[-b,b].$ Because the terms of the series are odd, it's enough to prove uniform convergence on $[0,b].$
Let $0<\epsilon< \pi/2.$ Choose $N$ such that $b/N < \epsilon.$ Note that the sequence $\sin (x/N), \sin (x/(N+1)), \sin (x/(N+2)),\dots$ is then decreasing for each $x\in [0,b].$
Suppose $N\le m <n.$ Claim:
$$\tag 1 |\sum_{k=m}^{n}(-1)^n\sin (x/k)| <\epsilon \text { for } x\in [0,b].$$
Proof: The terms in this series decrease in absolute value and are alternating in sign. Therefore $(1)$ is no larger than $|\sin(x/m)|\le |x/m| \le b/m\le b/N < \epsilon.$ This proves the claim.
Note that the claim says precisely that the partial sums of our series are uniformly Cauchy on $[0,b].$ Hence the series converges uniformly on $[0,b].$
