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

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.

• Could you please post your solution for the convergence of $\sum a_n \sin(nx) \implies n a_n\to 0$? Feb 25 '15 at 22:35

You can try to write :

$a_n*sin(n*x) = n*a_n*\frac{sin(n*x)}{n}$

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

with this and the condition n*$a_n$ -> 0, you can find something

$$\Longrightarrow$$

$$\forall\varepsilon>0,\;\exists N\in\mathbb{N},\;\forall n\geq N$$ 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 $$na_n\to0\,(n\to+\infty)$$.

$$\Longleftarrow$$

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.