TITLE: triangular series and inequality
QUESTION [5 upvotes]: Let $a_1,a_2,\ldots,a_n\in \mathbb R$ and $f(x)=a_1\sin x+a_2\sin 2x+\ldots+a_n\sin nx$  such that $|f(x)|\leq|\sin x|$ for every $x\in \mathbb R$. Prove that $|a_1+2a_2+3a_3+\ldots+na_n|\leq1$.
Solve: I solved it very simple.
$$|a_1+2a_2+\ldots+na_n|=\lim_{x\to 0}\frac{|a_1\sin x+a_2\sin 2x+\ldots+a_n\sin nx|}{|\sin x|}=\lim_{x\to 0}\frac{|f(x)|}{|\sin x|}\leq1.$$

REPLY [5 votes]: Observe that $|a_1 +2a_2 + \cdots + na_n| = |f^\prime(0)|$ and that $f(0)= 0$. 
Suppose for contradiction that $f^\prime(0) > 1$, say $f^\prime(0) = 1 + \varepsilon$ (the case when $f^\prime(0) < -1$ is the same).  Since $f^\prime(x)$ is continuous, there is some $\delta>0$ such that $f^\prime(x) > 1+\frac{\varepsilon}{2}$ for all $0\leq x \leq \delta$.
Apply the fundamental theorem of calculus:  $\displaystyle f(\delta) = \int_0^\delta f^\prime(x)\ dx \geq \int_0^\delta (1+\frac{\varepsilon}{2})\ dx = \delta(1+\frac{\varepsilon}{2}) > \delta > |\sin(\delta)|$, a contradiction.