$ \sum a_n \sin (nx)$ converges uniformly iff $na_n\to 0$ as $n \to \infty$

Question

Let $\{a_n\}$ be decreasing sequence of positive terms then prove that $\displaystyle \sum a_n \sin (nx)$ converges uniformly on $\Bbb{R}$ iff $na_n \to 0$ as $n\to \infty$.

I proved that convergence of $ \sum a_n \sin (nx)$ implies $na_n \to 0$ as $n\to \infty$ . I got stuck while prove the converse, I tried ussing Dirichlet's test but the problem here is that partial sums of $\displaystyle \sum^{n}_{k=1} \sin (kx)$ are bounded by $\displaystyle \frac{1}{|\sin (\frac{t}{2})|}$ , so as per Dirichlet's test requirement I'm not getting uniform bound.

Answer 1

If: $$ D_N(x) = \sum_{n=1}^{N}\sin(nx) $$ then we have: $$\left| D_N(x) \right|\leq \min\left(N,\frac{1}{|\sin(x/2)|}\right) $$ and the claim follows by partial summation.

Answer 2

$\Longrightarrow$

$\forall\varepsilon>0,\;\exists N\in\mathbb{N},\;\forall n\geq N$ and $\forall p\in\mathbb{N}$, we have $$ \Big|a_n\sin nx+a_{n+1}\sin(n+1)x+\cdots+a_{n+p}\sin(n+p)x\Big|<\varepsilon. $$

By letting $n>2N$, we have $$ \Big|a_{[\frac{n}{2}]}\sin [\frac{n}{2}]x+a_{[\frac{n}{2}]+1}\sin([\frac{n}{2}]+1)x+\cdots+a_{n}\sin(n)x\Big|<\varepsilon. $$

Since the convergence is uniform, we can take $x\in[\frac{\pi}{4[\frac{n}{2}]},\,\frac{3\pi}{8[\frac{n}{2}]}]$ to ensure $\sin kx\geq\frac{\sqrt{2}}{2}$ for $k=[\frac{n}{2}],\cdots,\,n$.

Then we have $\frac{\sqrt 2}{2}a_n(n-[\frac{n}{2}])<\varepsilon$, and from which it's easy to check that $na_n\to0\,(n\to+\infty)$.

$\Longleftarrow$

Answer 3

You can try to write :

$$a_n \sin (nx) = n a_n \frac{\sin(nx)}{n}$$

Prove that the series $\sum \frac{\sin(nx)}{n}$ converges, I think you can even have good information about its limit, by calculating the partial sum of the $\cos(nx)$ and integrating it.

with this and the condition $na_n \to 0$, you can find something