If $(c_n)$ is a decreasing sequence of positive real numbers and if $\sum_n c_n\sin{nx}$ is uniformly convergent, then $\lim{(nc_n)}=0$

Suppose there is $\epsilon > 0$ and $n_1,n_2,n_3,...$ going to infinity such that $n_k c_{n_k} > \epsilon $ for each $k$. We will show that the series cannot converge uniformly.

Since $\sin(x) > {1 \over 2} x$ whenever $0 < x < 1$, whenever $0 < x < {1 \over n_k}$ one has $$\sum_{{n_k \over 2} \leq n \leq n_k} c_n \sin{nx} > {x \over 2} \sum_{{n_k \over 2} \leq n \leq n_k} n c_n$$ Since the $c_n$ are decreasing, this is at least $$ {x \over 2} \sum_{{n_k \over 2} \leq n \leq n_k} n c_{n_k}$$ $$ \geq {x \over 2} \sum_{{n_k \over 2} \leq n \leq n_k} {n_k \over 2}c_{n_k}$$ $$ \geq {x \over 2}({n_k \over 2} - 1){n_k \over 2}c_{n_k}$$ We have ${n_k \over 2} - 1$ here in case $n_k$ is odd. So if $x = {1 \over 2n_k}$ for example, we therefore have $$\sum_{{n_k \over 2} \leq n \leq n_k} c_n \sin{nx} > {1 \over 4n_k}({n_k \over 2} - 1){n_k \over 2}c_{n_k}$$ $$\geq {1 \over 16} n_k c_{n_k}$$ $$\geq {\epsilon \over 16}$$ Thus the terms of the series from ${\displaystyle {n_k \over 2}}$ to ${\displaystyle n_k}$ sum to at least ${\displaystyle{\epsilon \over 16}}$ at one point. Thus if we choose a sequence of such ${\displaystyle n_k}$'s, call them ${\displaystyle n_{k_l}}$, such that ${\displaystyle n_{k_l} > 2 n_{k_{l-1}}}$ we see the sum cannot converge uniformly; the next bracket of terms from ${\displaystyle n = {n_{k_l} \over 2}}$ to ${\displaystyle n = n_{k_l}}$ will always contribute at least ${\displaystyle{\epsilon \over 16}}$ at some $x$.