Uniform convergence of $\sum n^\alpha \sin{\frac{x}{n}}$

Question

I want to find the range of $a \in \mathbb{R}$ such that $\displaystyle \sum_{n = 1}^\infty n^a \sin{\frac{x}{n}}$ converges uniformly on $[-\pi, \pi]$.

If $a < -1$, then I can prove it's uniform convergence with $\displaystyle \sum_{n = 1}^\infty n^a \Big|\sin{\frac{x}{n}}\Big| \geq \displaystyle \sum_{n = 1}^\infty n^a$.

Also, If $a \geq 1$, then I can say it does not uniformly converge. It is because

$n^{a}\Big|\sin{\frac{x}{n}}\Big| \geq n\Big|\sin{\frac{x}{n}}\Big| = |x|\Big| \frac{\sin{\frac{x}{n}}}{\frac{x}{n}}\Big| \rightarrow |x| \quad (n \to \infty)$

and If $x \neq 0$, then $\displaystyle \sum_{n = 1}^\infty n \sin{\frac{x}{n}}$ does not converge.

So, I should to consider weather it uniformly convergences or not at $a \in [-1,1)$. Please some idea.

Answer 1

Hint: In addition to @RobertIsrael's hint, one may proceed as follows: Using the fact that $t\mapsto\frac{\sin t}{t}$ decreases on $[0,\pi/2]$, one gets that for all $0<x\leq M$ and $n\geq M$ $$ 0<\sin1 \frac{x}{n^{1-a}}\leq n^a\sin\tfrac{x}{n}\leq \frac{x}{n^{1-a}} $$ Divergence of the series follows for $a\geq0$. Uniform convergence follows for $a<0$.

Answer 2

Hint: by Taylor's theorem with Cauchy's form of the remainder $$ \left|\sin\left(t \right) - t \right| \le t^2/2 $$