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

This problem is from the book Introduction to Real Analysis by Bartle and Sherbert. I attempted it several times but never could get anywhere near the solution. Any help? Thanks and regards.
 A: 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 $x$.
A: Here is a sketch:


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*$c_n \to c$. Prove that $c = 0$. For this, choose $p$ very high, $x = 1/p$. By uniform continuity,$$f(x) - \varepsilon < c_1\mbox{sin }x + .... + c_p\mbox{sin }px < f(x) + \varepsilon$$ Now all the terms are positive, so $$c_1\mbox{sin }x + .... + c_p\mbox{sin }px \geq c(\mbox{sin }x + .... + \mbox{sin }px)$$ making $f(x)$ blow up at $0$, unless $c$ itself is $0$.

*Next rewrite your series as $$\sum \bigg[a_1\frac{\mbox{sin }x}{1} + .... + a_p\frac{\mbox{sin }px}{p} + ....\bigg]$$. WLOG (descending to a subsequence if need be) you can assume that $|a_p| > \delta$ for all $p$. Then, $$ a_1\frac{\mbox{sin }x}{1} + .... + a_p\frac{\mbox{sin }px}{p} \geq \delta\mbox{sin }x[1 + \frac{1}{2} + ... + \frac{1}{p}]$$, which still blows up at $0$ because of $\lim_{y \to 0}\frac{sin y}{y} = 1$.
