prove that lim($nc_n$) = $0$

Question

If {$c_n$} is a decreasing sequence of positive numbers. If $\sum (c_n\sin\ nx)$ is uniformly convergent.

Prove that lim($nc_n$) = $0$

I do not know how to begin this prove, all I have made used of is that the fact that series is uniformly convergent, that means the sup-norm must $\rightarrow 0$, I've yet made of of the fact the $c_n$ is positive decreasing sequence.

Any insight or help is deeply appreciated.

Answer 1

We use the fact that for $0 \leqslant x \leqslant \frac{\pi}{2}$ we have $\sin x \geqslant \frac{2}{\pi}x$. Let $\varepsilon > 0$. By the uniform convergence of the series, there is a $K_\varepsilon \in \mathbb{N}$ such that for all $K_\varepsilon \leqslant M \leqslant N$ we have

$$\Biggl\lvert\sum_{n = M}^N c_n \sin (nx)\Biggr\rvert \leqslant \frac{\varepsilon}{2}$$

for all $x\in \mathbb{R}$. In particular, for $x = \frac{\pi}{2N}$, we have

\begin{align} \varepsilon &\geqslant 2\Biggl\lvert \sum_{n = K_\varepsilon}^N c_n \sin \frac{n\pi}{2n}\Biggr\rvert \\ &= 2\sum_{n = K_\varepsilon}^N c_n \sin \frac{n\pi}{2N} \\ &\geqslant \frac{2}{N}\sum_{n = K_\varepsilon}^N n\cdot c_n \\ &\geqslant \frac{2}{N}\sum_{n = K_\varepsilon}^N n\cdot c_N \\ &= \frac{2c_N}{N} \sum_{n = K_\varepsilon}^N n \\ &= \frac{c_N}{N}\bigl(N(N+1) - K_\varepsilon(K_\varepsilon - 1)\bigr) \\ &= Nc_N + c_N\biggl( 1 - \frac{K_\varepsilon(K_\varepsilon - 1)}{N}\biggr). \end{align}

It follows that for $N \geqslant K_\varepsilon(K_\varepsilon - 1)$ we have $N\cdot c_N \leqslant \varepsilon$.