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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.

Mittens
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jamize
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2 Answers2

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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 $a<0$.

Mittens
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  • I understand uniform convergence follows for $a < 0$ by the inequality and M-test. Also, I cannot confirm how to use the inequality to prove divergence follows for $a \geq 0$, and I am thinking about it... – jamize Jan 22 '21 at 20:21
  • Thank you for your hint and kindness! – jamize Jan 22 '21 at 22:28
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Hint: by Taylor's theorem with Cauchy's form of the remainder $$ \left|\sin\left(t \right) - t \right| \le t^2/2 $$

Robert Israel
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